mm-387

===
Subject:
: Re: chinese remainder theorem> Have a look at
my method for the chinese remainder theorem.> It's at
http://www.geocities.com/dirkie6/chinese.html> I would like
some comments on the originality of this method.It's required
to find x satisfying x = a_i mod m_i, i = 1, 2, ..., n, and
we'll assume gcd(m_i, m_j) = 1 for i < j. Let M be the product
of all the m_i. Let M_i = M / m_i, i = 1, 2, ..., n. Let r_i be
the inverse of M_i mod m_i. Then x = r_1 a_1 M_1 + ... + r_n
a_n M_n. This is standard. It's in many intro number theory
texts. So far as I can see, what you've done is you've
implemented this formula, only you've restricted it to the m_i
being prime numbers, and you're doing it 2 moduli at a time
instead of all at once.-- Gerry Myerson
(gerry@maths.mq.edi.ai) (i -> u for email)===
===
Subject:
: Hex Win
Proof?It is an old theorem that in Hex, once the board has
been completelyfilled in with two colours, there *must* be a
winning path for oneor other of them.Now, I can prove this
easily enough mathematically, but I'm wondering ifthere is a
simple proof, or proof outline, that would be
understandableand reasonably convincing to the intelligent
layman.Can anyone help out please? ===
===
Subject:
: Re: Hex Win
Proof?> It is an old theorem that in Hex, once the board has
been completely> filled in with two colours, there *must* be a
winning path for one> or other of them.> Now, I can prove this
easily enough mathematically, but I'm wondering if> there is a
simple proof, or proof outline, that would be understandable>
and reasonably convincing to the intelligent layman.> Can
anyone help out please?Looks like an opportunity for an
inductive proof. Show that there is alwaysa winner for a small
board, where the number of cases is small and all canbe drawn
quickly. Then show that adding a row and column to a
finishedboard will always yield a winning path for one player
or the other. ===
===
Subject:
: Re: Hex Win Proof?> It is an old
theorem that in Hex, once the board has been completely>
filled in with two colours, there *must* be a winning path for
one> or other of them.> Now, I can prove this easily enough
mathematically, but I'm wondering if> there is a simple proof,
or proof outline, that would be understandable> and reasonably
convincing to the intelligent layman.> Can anyone help out
please?Imagine the board tipped up, so player A goes from top
to bottom,player B from left to right. Suppose the board is
actually two sheetsof clear plastic, and two players have
stones of white sugar (A) andred granite (B), and moreover
these stones plug the hexagon cells(in the obvious way).Now
pour water in at the top. Either it dissolves through the
sugarand reaches the bottom, in which case A has a path, or it
doesn't, inwhich case there must be a continuous boundary of
impervious granite,and B has a path.(Does this help? I think
it does (slightly) use some facts about thebehaviour of
liquids to help the imagination
along.)http://imaginatorium.org===
===
Subject:
: Re: Hex Win
Proof?>It is an old theorem that in Hex, once the board has
been completely>filled in with two colours, there *must* be a
winning path for one>or other of them.>Now, I can prove this
easily enough mathematically, but I'm wondering if>there is a
simple proof, or proof outline, that would be
understandable>and reasonably convincing to the intelligent
layman.>Can anyone help out please?Neither of the proofs
(which are basically the same) posted so far iscorrect. Both
would apparently conclude that a winning path would beformed
on a squared board, whereas this is not the case - a
squaredboard could end in a draw.An actual proof must use the
hex nature of the board or,alternatively, that 3 cells meet at
each vertex. A proof is given inCameron Browne's book Hex
Strategy, but whether it would convince anintelligent layman
is not clear.Maybe a simpler proof could be achieved by
induction?Jonathan Welton===
===
Subject:
: Re: Hex Win
Proof?Mime-version: 1.0Content-type: text/plain;
charset=US-ASCIIContent-transfer-encoding: 7bit> ... A proof
is given in Cameron Browne's book Hex Strategy, but whether
it> would convince an intelligent layman is not clear.> Maybe
a simpler proof could be achieved by induction?My experience
with intelligent laymen is that they don't accept
induction.Less intelligent laymen accept it, but intelligent
laymen think you are justusing circular reasoning.===
===
Subject:
:
Re: Hex Win Proof?> Neither of the proofs (which are basically
the same) posted so far is> correct. Both would apparently
conclude that a winning path would be> formed on a squared
board, whereas this is not the case - a squared> board could
end in a draw.> An actual proof must use the hex nature of the
board or,> alternatively, that 3 cells meet at each vertex. A
proof is given in> Cameron Browne's book Hex Strategy, but
whether it would convince an> intelligent layman is not
clear.> Maybe a simpler proof could be achieved by induction?>
Jonathan WeltonI wasn't assuming a square board, I was
imagining the board set up like a parallelogram. At least,
that is how I orientate the board when I play. Then red goes
top to bottom and blue goes left to right (red and blue
because the board I made uses poker chips).What would be more
interesting is trying to explain to a lay person that whoever
goes first should win, unless they screw it up. That is why
whenever I play, I always go second. If I lose, it was
destined. -- ===
===
Subject:
: Re: Hex Win Proof?>Neither of the
proofs (which are basically the same) posted so far
is>correct. Both would apparently conclude that a winning path
would be>formed on a squared board, whereas this is not the
case - a squared>board could end in a draw.>An actual proof
must use the hex nature of the board or,>alternatively, that 3
cells meet at each vertex. A proof is given in>Cameron Browne's
book Hex Strategy, but whether it would convince an>intelligent
layman is not clear.>Maybe a simpler proof could be achieved by
induction?>Jonathan Welton> I wasn't assuming a square board, I
was imagining the board set up like> a parallelogram. At least,
that is how I orientate the board when I> play. Then red goes
top to bottom and blue goes left to right (red and> blue
because the board I made uses poker chips).> What would be
more interesting is trying to explain to a lay person that>
whoever goes first should win, unless they screw it up. That
is why> whenever I play, I always go second. If I lose, it was
destined.If you are a really good player, you go second, and
tell your opponent themoment he loses the
advantage.===
===
Subject:
: Re: Hex Win Proof?> It is an old theorem
that in Hex, once the board has been completely> filled in with
two colours, there *must* be a winning path for one> or other
of them.> Now, I can prove this easily enough mathematically,
but I'm wondering> if there is a simple proof, or proof
outline, that would be> understandable and reasonably
convincing to the intelligent layman.> Can anyone help out
please?Here's what I'm thinking...Suppose you have red going
top-bottom and blue going left-right. If red does not win,
then there must be a path that divides the board into to
pieces, a top and a bottom piece. Red cannot be in any
position in that path, so it must be blue. Thus blue wins.
Rotate pi/2 and switch colors. -- ===
===
Subject:
: very easy
analysis problem.......lim nx{(1+x)^n}{(1-x)^n} =
0n->00(0<x<1)------------------------um........i can't
understand.sorry.....let me explain. please===
===
Subject:
: Re:
very easy analysis problem.......> lim nx{(1+x)^n}{(1-x)^n} =
0> n->00> (0<x<1){(1+x)^n}{(1-x)^n} = (1-x^2)^n. Now use the
fact that if 0 <= a < 1, then n*a^n -> 0.===
===
Subject:
: Re:
Ability In Mathematics>Did I just read that they are relaxing
the proposed requirement that>math teachers in high school
must have degrees in math? (and science,>too) It seems there
is a shortage of people with math/science>degrees willing to
work at teachers' wages/conditions.They did, sort of. In the
US, the decisions about what it takes to be a
secondary-schoolmathematics teacher are left to the states.
But nationwide there has beenconcern about student
performance. One proposed solution was theNo Child Left Behind
(NCLB) Act which tried to mandate all sorts ofimprovements
which the states would have to carry out (without
anyadditional resources coming from the federal government).
Not surprisingly,the mere passage of the Act did not suddenly
create more qualifiedmath (and science) teachers, and so the
current US administration hasbeen backpedalling on that
particular requirement.In my state (Illinois) it does take
something comparable to a mathdegree to be certified as a HS
math teacher. But -- teachers are allowed toteach a certain
number of courses in areas in which they are notcertified, and
some school districts may hire non-certified teachersbecause of
extenuating circumstances. (Also, private schools can
hireanyone they want to teach.) So in practice there are
plenty of high school students who are learning mathematics
from teacherswho themselves have had nothing more than a few
semesters of calculus,if that. And then some of those high
school students show up at the university,wondering why
they're not ready for regular university math courses.
Sigh.dave===
===
Subject:
: Re: Ability In Mathematics> (without any>
additional resources coming from the federal
government)Actually, 50 percent more than under Clinton. But
still inadequate, ofcourse.===
===
Subject:
: Re: Ability In
Mathematics>You could have been describing my wife with this
statement. She was >told in high school by a teacher that she
would NEVER be able to succeed >in the sciences due to her
ability in math. It turns out that she had >teachers who
simply didn't care enough to explain it to her. Later, she
>met a teacher who explained the math to her in clear
terms.Married him, did she?===
===
Subject:
: Re: probability of the
linear combination of inequalities> If for any vector A and
its approximation A' we have > Pr{|A'|^2 >= (1+c) |A|^2} <= p
(1)Ummm, before you go applying that to all sorts of
vectors,how would that work when A = 0?Department of
Mathematics http://www.math.ubc.ca/~israelUniversity of
British ColumbiaVancouver, BC, Canada V6T 1Z2===
===
Subject:
: Re:
Hypocritical Fortune Tellers (What are the odds?)> [...] The
mathematicians> have tools to analyse sets of numbers but
their tools> will not for example, show us that the 29x29th
chapter> of the Bible contains 29 verses. The mathematicans>
are too closed minded to stop and look at some of> the
patterns, and here one of them simply doesn't> want the topic
discussed.One fact does not a pattern make. If the (K*K)th
chapter of the Biblecontained K verses, for K = 1, 2, 3, etc.
up to floor (sqrt(C)) (whereC is the number of chapters in the
Bible, a number I don't know rightoff, but you probably do),
THAT would be a pattern. An example of a real pattern is the
following: (3 + 5)/2 = 4, theinteger between 3 and 5; (7 +
9)/2 = 8, the integer between 7 and 9;(100 + 102)/2 = 101, the
integer between 100 and 102. This can be provento work for any
two numbers which are two apart: Let n be any integer; then(n
+ (n+2))/2 = (2n + 2)/2 = 2(n+1)/2 = n+1, the integer between
n and n+2. Here's a couple of facts: 14 (base 5) is 9 (base
10), and 1, 4, and 9 are perfect squares. 49 (base 10) is a
perfect square, as well as its digits,4 and 9. Pattern or
coincidence? For Numerology (I'm capitalizing it out of
respect, for the moment,until and unless I reject it), if you
could show that every fact about every person in every one of
your posts was correct (personality traits,etc), and could do
so even if the people were chosen at random (not justwhat you
select to post), then Numerology would pass a test as being
atrue science. A related test might be: I give you five names
and birthdates. Youare to determine which (if any) are fakes.
-- Christopher Heckman===
===
Subject:
: Re: Hypocritical Fortune
Tellers (What are the odds?)>[...] The mathematicians>have
tools to analyse sets of numbers but their tools>will not for
example, show us that the 29x29th chapter>of the Bible
contains 29 verses. The mathematicans>are too closed minded to
stop and look at some of>the patterns, and here one of them
simply doesn't>want the topic discussed.> One fact does not a
pattern make. If the (K*K)th chapter of the Bible> contained K
verses, for K = 1, 2, 3, etc. up to floor (sqrt(C)) (where> C
is the number of chapters in the Bible, a number I don't know
right> off, but you probably do), THAT would be a pattern.> An
example of a real pattern is the following: (3 + 5)/2 = 4, the>
integer between 3 and 5; (7 + 9)/2 = 8, the integer between 7
and 9;> (100 + 102)/2 = 101, the integer between 100 and 102.
This can be proven> to work for any two numbers which are two
apart: Let n be any integer;then> (n + (n+2))/2 = (2n + 2)/2 =
2(n+1)/2 = n+1, the integer between n andn+2.> Here's a couple
of facts: 14 (base 5) is 9 (base 10), and 1, 4, and 9> are
perfect squares. 49 (base 10) is a perfect square, as well as
itsdigits,> 4 and 9. Pattern or coincidence?> For Numerology
(I'm capitalizing it out of respect, for the moment,> until
and unless I reject it), if you could show that every fact
about> every person in every one of your posts was correct
(personality traits,> etc), and could do so even if the people
were chosen at random (not just> what you select to post), then
Numerology would pass a test as being a> true science.> A
related test might be: I give you five names and birthdates.
You> are to determine which (if any) are fakes.> --
Christopher HeckmanIf you are an asshole and you change your
name, you willstill be an asshole. The names people have and
the birthdayspeople have do not influence their personalities.
The namesare in harmony with the birthdays and with the numbers
inthe Bible, this is revealing something about God's
personality(and great power), and not of the individual's
personality.You are denigrating my work based upon flaws in
yourassumptions of what I am doink and trying to show withmy
work -Daryl S. Kabatoff===
===
Subject:
: Re: Hypocritical Fortune
Tellers (What are the odds?)Daryl S. Kabatoff
<darylsk@shaw.ca> stumbled drunkenly into thegroup and rudely
farted the following noxious cloud:> A) The work I do with
math has nothing to do with the Bible CodeThe work
(Translation: spamming) you do scumbag has nothing to dowith
math or the Bible. It's just the way you kill time while
sittingon your fat duff 24/7 trying to get attention.One wants
to be loved; failing this, to be admired; failing this,
tobefeared; failing even this, to be hated and despised. One
wants toarousesome sort of feeling in people. The soul shrinks
from the void andwantscontact at any price. ~ Hjalmar
Soderberg> B) The Jews were very diciplined, their tents were
perfectly set> into rows, they wore clean garments, they
numbered their soldier> accurately.They had JOBS scumbag. Now
there's a novel concept for you. Howabout getting one instead
of embarrassing yourself here begging formoney or assistance
and pestering these poor people.> C) There are many different
ways to give testimony of God's power,> why don't you continue
with the many excercizes you have lined up> without trying to
denigrate my work?Because scumbag, what you do isn't work,
it's just annoyingoff-topic spamming from an unemployed,
stalker, pervert and pedophile. You should make everyone happy
and safer by offing yourself. Soon,OK? Or at
least....................BEAT IT.Some cause happiness wherever
they go; others whenever they go. ~Oscar Wilde> -Daryl S.
Kabatoff-Documented jag-off and professional LOSER.<<<snipped
a lot of psycho crap from Shawn pedophile-boy Kabatoff>>The
following (courtesy of Waxy.org) is sort of an unofficial
FAQexplaining the psychotic nonsense posted to Usenet by Shawn
DarylKabatoff AKA Dar, AKA Probababbilities. And now AKA marcia
andme.WARNING: Read below before even thinking about responding
to
thistwit.http://www.waxy.org/archive/2002/05/21/dar_kaba.shtml
#000643Usenet has the tendency to provide a public forum for
those who wouldnormally be scribbling in a closet. For
example, take Daryl ShawnKabatoff. For the last few years,
he's methodically gatheredstatistics from various sources,
ranging from local newspaperobituary pages to the food court
of the Saskatoon Midtown Plaza mall.With all the raw data he's
collected, he's attempting to prove dailythat our full names
are in mathematical harmony with our birthdays.His rants
normally focus on a single individual he's met or readabout,
starting with calculations related to their birthdate and
fullnames, blending in whatever other personal information
about theirfamily members, spouses, birthplace, and career
he's been able tozealotry, and personal torment. I've never
seen anything like it.With all the prime numbers, Fibonacci
sequences and biblicalreferences, it's like reading the
notebooks of Maximillian Cohen andJohn Nash combined.
Unsurprisingly, several posts unfold to reveal ahistory of
painful mental illness. If you have some time, take a
look.I've detailed his posting history and a several sample
posts below. Usenet Posting History:January 27, 1999 to July
5, 2000 as Catsco@home.comDecember 9, 2000 to May 4, 2001 as
s.kabatoff@sk.sympatico.caOct 30, 2001 to Oct 31, 2001 as
kabatoff@the.link.caJanuary 20, 2002 to April 17, 2002 as
s_kabatoff@hotmail.com (originalposts have been removed from
Google Groups archive)April 26, 2002 to Present as
dar_kabatoff@hotmail.comSelected Posts:Tessa Lynne
SmithDastageer Sakhizai and Helen SmithBrett David MakiAndrew
Meredith CottonKathryn Lee HippersonAmanda Dawn NewtonMona
Marie EtcheverryTony Peter NusplLisa Charlene McMillanGrant
Allyn Wood Comments scarier still is that saskatoon is my
hometown, though not my currentresidence. and every single
place he's mentioned in his posts (mostnotably nervous
harold's and the roastary) were either places i'vebeen (as
it's a small city of 200K) or hangouts, ie. the two
placesmentioned. chances are i could email some friends back
home and findout if they know of him, they (my friends that
is) being of thebroadway-centred slacker ilk. myself, too,
until i got out of there.eh, anyways. thought it odd to see
all this. midtown mall. i ate mymeals there, whilst waiting
several days in line for star wars episodeone, at the theatre
across the street.posted by andy raad on May 22, 2002 06:20
PMFascinating. It's like he's trying to take chaos and bind it
intowhatever rules he can find, religious, logical and
otherwise. Numbersand math have a reliable pattern, something
that can always be provento true or false. People and religion
do not. It reminds me of DarrenAronofsky's movie Pi. It's the
story of an paraniod genius who istrying to find a pattern in
Pi. A group that takes interest in hiswork is convinced that
the existence of Pi, a number whose existencecan be proven but
no quantified, is proof of the existence of God.Kabatoff's hunt
for patterns in something as random as name selectionis a way
to reconcile his deeply logical thought process with
hisconflicting religious views.Exactly. I probably shouldn't
have, but I e-mailed Daryl yesterday,asking him if he'd be
willing to create a numerological analysis forme. I also asked
him if he had seen either Pi or A Beautiful Mind, andwhat he
thought of them. If he replies, I'll be sure to post it.I
baked many pumpkin pies for Shawn (he likes pumpkin pies). I
rubbedpumpkin pie all over my breasts for him, and my breasts
turned orange.I am a pumpkin for Shawn.posted by Trisha
Blondie on July 24, 2002 10:41 PMUm, that's swell. So, you're
in love with him?Shawn once went to a funeral for a Jehovah
Witness that shot himselfand the lemon tarts were very bad,
they were not only sour but wererubbery as well. Shawn said
that the guy was some kind of JehovahWitness prophet, he saw
in advance that the lemon tarts at his funeralwere to be very
very bad, and so he shot himself. Shawn said that henever ate
pumpkin pie at a funeral but would like to some day.
Shawnlikes pumpkin pie and so I have been practicing to make
very goodpumpkin pies.posted by Trisha Blondie on July 25,
2002 02:49 PMShawn said that the lemon tarts were sour, bitter
and rubbery.I don't think this guy takes notes. I think he has
Total Recall, andit has driven him insane...Oh... I almost
forgot... I didnt spend thousands of dollars a daytormenting
Daryl... We got a deal on tormenting that fiscal year, itonly
came to about 37cents a day....Mr. Kabatoff attempts to
portray himself as a victim, but in fact heis a violent
predatory pedophile who is well known to his local
lawenforcement. In his post to multiple newsgroups with the
subjectCollecting Mail For The Coming Anti-Christ, he
encourages mothers tosend him photos of their naked daughters.
Mr Kabatoff explains, Ipersonally did not want photographs
being mailed to (the comingAnt-Christ) that were of underage
children unless the parent wassigning consent. He is banned
from virtually all the shopping mallsin his community because
he stalks young people and sexually harassesthem. He has an
extensive arrest record which includes sexualmolestation
charges. He's been hospitalized in mental institutionsabout
his contact with young girls in many posts. Search
newsgrouparchives for posts by him containing the word nubile.
As part of hisharrassment, he provides personal details in a
public forum, such asthe real names of real children, in these
and other posts. About onewanted her and her sister
dead.http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+
dead+or+in+my+bed&hl=en&lr=&ie=UTF-8&selm=asqm35%24tjq5j%241%
40ID-136124.news.dfncis.de&rnuHe not only curses children and
prays for their death in his posts, healso enjoys attending
the funerals of young people: And so, sincenubile sweeties are
found in greatest abunce at the funerals ofhigh school
students, then it is the funerals of high school studentsthat
make the very very best funerals, especially if there is
food...I stuff my face (and my pockets) with all the good food
and look atall the pretty nubile sweeties and have the time of
my
life...http://groups.google.com/groups?q=Daryl+Shawn+Kabatoff+
nubile+sex&hl=en&lr=&ie=UTF-8&scoring=d&selm=LfXN8.63042%
24R53.25142039%40twister.socal.rr.com&rnum=1 Many of his posts
are sent to alt.teens.advice. However, he liberallyspams,
floods and crossposts his off-topic threatening and
offensivemissives to countless newsgroups. Some people HAVE
problems and somefolks ARE problems. Don't dismiss Mr.
Kabatoff as a harmless nut. Whenhe sends these posts to any
newgroup, please help by reporting him toI knew of him when I
was attending the University of Saskatchewan.He'd hang out in
the Arts computer lab and all you'd see is screens ofnumbers
racing by on his laptop. I have an original copy of
hisCollecting Mail for the Coming Anti-Christ pamphlet, and
have seenhim be hauled away by campus security on more than
one occasion. Myfriends and I refer to him as Crazy Number
Man.I've been posting to (and about) Shawn for over two years
with biggaps in between. He has seen Pi and didn't like it and
didn't think itresembled him at all. (Wrong, it fits him to a
tee) He doesn't havetotal recall and has stated that he
travels with a lap top to notateitems. Also, he uses cut n'
paste a lot if you read all the waythrough his ramblings. He
is anti-social as shown by his angrystatements towards those
who, by his own admission, have been kind(but not kind enough)
to him. Still, he's intelligent and seems to beable to take a
joke on occassion. That's where I came in. ALOHAReply to
group(Unsolicited e-mail is deleted from the server unread if
it comes from anyone not already in my addressbook. I'll never
even see it)===
===
Subject:
: Re: Antidiagonal, InfinityI look to the
unit impulse function to consider its definition andthus
meaning within its context.The unit impulse function, Dirac's
Delta, differs from Kronecker'sDelta, Kronecker's Delta f(i,
j)=1 when i=j, f(i, j)=0 when i =/= j,and applied
componentwise (?) to a matrix leaves its diagonal. Dirac's
Delta, the unit impulse function, equals infinity for f(0),and
zero for f(x) for x =/= 0.One part of the definition of Dirac's
Delta is that integrated overthe reals its evaluation is equal
to one, the same as that of the unitsquare, perhaps the most
simple description of a closed shape witharea one, a square of
side length one.Consider the unit square divided in half by the
y axis, with its sidesdefined by f(x)= {0,1}, f(y)={-1/2, 1/2},
defined for x on [-1/2, 1/2]and for y on [0,1]. Another
quadrilateral with the same area haswidth 1/2 and height 2, as
does the rectangle with width 1/4 andheight 4, etcetera, for
width 1/x and height x. The value of 1/x * xis always one.The
function (the unit impulse function) is not defined to be
anythingother than zero except at zero, 1/x * x = 1.Zero is a
point, an element of the set of reals on the real numberline,
and only one point.It's not said to be defined as anything
other than zero for any valueother than zero.One concept of an
integral is that its evaluation's result is the sameas the area
under the curve of the graph or plot of the function ofone
variable, where when the function is less than zero
itscontribution to the integral is negative. The evaluation of
theintegral is exactly equal to a geometric determination of
the area ofa shape bounded by the variable's axis above the
axis. The geometricintuition exactly represents a result of
what Newton and Leibnizcalled the infinitesimal calculus: the
integral calculus, the sum ofinfinitesimals. In later years in
a fight for a more rigorousfoundation under the beliefs of the
time, some hundred years beforeRobinson's non-standard
analysis, allusions to infinitesimals werereplaced with finite
constructions.Let us consider an alternate geometric analysis
of the value of theunit impulse function. Start as above with
a unit square centeredabout the y axis. Chop off its sides, as
above, but instead ofreforming a rectangle of half the area
immediately above it, move theremoved area to be centered
about x=1. Again doubling, we can seethat each rectangle has
unit height and its width is 1/2^x, thevariable there being
the number of iterations, and there are then 2^xmany (as many
as x) of those rectangles, each of height equal to one. The
sum of their area is a constant, and equal to one. Repeat
untileach rectangle is point width and there are infinitely
many of them. Why or why not does the area change?What is the
area?I don't very well understand the unit impulse function.
I've seen itdescribed as having periodic aspects about zero
and not necessarilydefined to be zero for a non-zero variable.
Hopefully one of thelocal sci.math denizens with knowledge of
the unit impulse function,for example many signals engineers,
can help to explain it, and alsothe geometric consideration as
described above. The impulse functioncan be scaled, by a
scalar, I'd appreciate some explanation of that.Have I put
forth enough information to explain to you why I think itsarea
is one, and/or two, and why? Why or why not?Warm regards, Ross
F.===
===
Subject:
: Re: Antidiagonal, Infinity> I look to the unit
impulse function to consider its definition and> thus meaning
within its context.> Dirac's Delta, the unit impulse function,
equals infinity for f(0),> and zero for f(x) for x =/= 0.The
Dirac Delta is not a real function.===
===
Subject:
: Angle
Subtending Arc and Chordunknown angle x subtends known arc A
and chord BBx/(2A)=sin(x/2) z=(x/2)^2solve,(z^4)/9! - (z^3)/7!
+ (z^2)/5! - z/3! + (1-B/A)=0x =
2{2{2{2{c[4]}^(3/4)-c[3]}^(2/3)-c[2]}^(1/2)-c[1]}^(1/2) +
e-2c[q]=(1-B/A)/[(2q+1)!|N|^2] q=1,2,3,4 **DATA** ANGLES B/A
TRUE CALCULATED1 0 0.7182818280.964 0.939 1.0551638870.927
1.336 1.3275861970.891 1.645 1.6885045560.855 1.911
1.9672154150.818 2.149 2.2106618890.782 2.369 2.4206769070.746
2.576 2.6117203010.709 2.772 2.7929549250.673 2.96
2.9575585840.63672 3.14159 3.11423727===
===
Subject:
: Re: Angle
Subtending Arc and Chord> unknown angle x subtends known arc A
and chord B> Bx/(2A)=sin(x/2) z=(x/2)^2 .. solve.. For better
accuracy than truncated sine series,solve using iterationslike
Newton-Raphson.===
===
Subject:
: Re: Packing squares into a 1 x
Zeta[2] rectangle> addRectangle(t);> t.check();Calling check
after addRectangle does not seem right.-- Clive
Toothhttp://www.clivetooth.dk===
===
Subject:
: Re: Packing squares
into a 1 x Zeta[2] rectangle>> addRectangle(t);>> t.check();>
Calling check after addRectangle does not seem right.You're
quite right.But amazingly, correcting this does not have any
effect.I guess that adding the rectangle to the listjust uses
a pointer to the rectangle,so correcting its orientation with
t.check() still works.In any case, my program still fails when
adding square 12867 (with pi = 103993/33102)or square 12909
(with pi = 355/113).I wonder if it would be difficult for your
programto print out where in the list the new rectangles are
installed?That might be the simplest way to see where we
diverge.-- Timothy Murphy e-mail (<80k only): tim /at/
birdsnest.maths.tcd.ietel: +353-86-2336090,
+353-1-2842366s-mail: School of Mathematics, Trinity College,
Dublin 2, Ireland===
===
Subject:
: Re: Packing squares into a 1 x
Zeta[2] rectangle> addRectangle(t);> t.check();>Calling check
after addRectangle does not seem right.> You're quite right.>
But amazingly, correcting this does not have any effect.> I
guess that adding the rectangle to the list> just uses a
pointer to the rectangle,> so correcting its orientation with
t.check() still works.> In any case, my program still fails>
when adding square 12867 (with pi = 103993/33102)> or square
12909 (with pi = 355/113).> I wonder if it would be difficult
for your program> to print out where in the list the new
rectangles are installed?> That might be the simplest way to
see where we diverge.Timothy,I have seen your program failing
on square 12,867 and I have figured outwhat is happening.I got
my program and your program to each create a file of the
rectanglescreated and consumed. I set epsilon to zero in both
programs so as not tomuddy the water.It turns out that the
programs diverge on handling the square 1/2838. Andthis is
because the denominator n in your rectangle has reached about
308(decimal) digits, and 10^308 is about the limit for the
Java double type.So, java considers that the denominator to be
infinite and your x gets setto zero, and two rectangles are
lost. A few thousand squares later we runout of room! Here is
the relevant part of my trace of your program... (A_Bis the
fraction A/B)2838: [using rectangle with sides...]
16094981_32872272120[and...]
171997530384560983181872925319003522657602600519958814525499564
700221967241722679887629709877653379316842024689191762371826756
590393041246582442746971002255746092314809955977118592842113269
630387401896651673445535631307865594273049312658190311576753383
1332414823315447720602850206552080218215282151849_
374409414423608354377363831484362771526931857169620500833386773
853568734386518087654023735566636522967804885653344877644045700
188571956459824962018732031251163631772196993506082072776028652
605002147960661805426860523410214651102431673960072331313839768
96849177436921199857403270945301876141245749904000[candidate
new
rectangle...]w=
175947603122073175454540464881028009017663905279321263218060391
948147016721889576913353792630535832136881097253014797498559115
155949287200944743381428523085698305483834060618694091700591830
385800065810530688841338953555087567116754396188010455211153410
129267975708142017917035340747392440131724587450527069600h=
163815145554186460641767160553415616736205983274067383989410605
157265439789110291630889091803773640326818289535917065769825276
657160146158122364581820404939565525219159870811677679201617266
360399212153407004688917069976459523472898590211927515121769342
03901930577226304863128942485410210319985028795726222050502n=
359354104779033352204277771506561961009488720803320809972141508
133084820879760251527319947329234156623993313007656667240509978
937706183423856711947524968510542309934226172841228438941188449
585054144968401836895833650652888618923546062135678840820384128
246890212164872646483689623784284583650074472505960887392000x=
0.0, y=NaNDiscarding 0.0[other candidate new
rectangle...]w=
493252969983827765101155982472852461531506139462272498381655687
863835139453708131616554320494103364977009305836502213038762473
132057412446879738438932980996016494111680377711858327361138707
011439188872037554742375967358879128096557132649382773322301795
98298908701492036774086213268786173870246619689441485600h=
126622306123690398944424866633742762864513291333094013379894823
161763502776518763751698360581125495639180166669364576194682867
842743545956256769537535224986096656072666022847508258964477959
684656146923326933367101356819199654307098682923072177878923230
530969067006649981142949127478606266261477967761085584000n=
359354104779033352204277771506561961009488720803320809972141508
133084820879760251527319947329234156623993313007656667240509978
937706183423856711947524968510542309934226172841228438941188449
585054144968401836895833650652888618923546062135678840820384128
246890212164872646483689623784284583650074472505960887392000x=
0.0, y=0.0Discarding 0.0You could try this program
fragment:=====================================BigInteger
i=newBigInteger(
175947603122073175454540464881028009017663905279321263218060391
948147016721889576913353792630535832136881097253014797498559115
155949287200944743381428523085698305483834060618694091700591830
385800065810530688841338953555087567116754396188010455211153410
129267975708142017917035340747392440131724587450527069600);
BigInteger
j=newBigInteger(
359354104779033352204277771506561961009488720803320809972141508
133084820879760251527319947329234156623993313007656667240509978
937706183423856711947524968510542309934226172841228438941188449
585054144968401836895833650652888618923546062135678840820384128
246890212164872646483689623784284583650074472505960887392000);
double di = i.doubleValue();double dj = j.doubleValue();double
dk =
di/dj;System.err.println(i=<+i+);System.err.println(j=<+j+);
System.err.println(di=+di);System.err.println(dj=+dj);
System.err.println(i/j=+dk);==================================
===One other thing about your program that I noticed is that
the methodcheckOrder does not update lastx each time around
the loop so it willprobably never find any mis-ordering.--
Clive Toothhttp://www.clivetooth.dk===
===
Subject:
: Re: Packing
squares into a 1 x Zeta[2] rectangle>Using my>Mathematica
program I can pack 100,000 squares into a 1
by>(103993/33102)^2/6 rectangle in about 3 minutes. [As
opposed to about 12>minutes when using the exact value for
pi.]> ... and I can pack 1,000,000 squares into a 1 by
(103993/33102)^2/6> rectangle in about 3.5 hours. [As opposed
to about 11 hours when using the> exact value for pi.]Why not
retain the advantages of both the rational approximations
andthe exact computation by using them as bounding functions,
in thefollowing way -lets say you had written pseudocode
something like this:if(f <= g)then {do what you do when it
fits}else {do what you do when it doesn't}but you also had
fast-to-compare bounds b1 < g < b2if(f <= b1)then !fast
comparison {it definitely fits - do what you do when it
fits}else if(f >= b2)then !fast comparison {it definitely
doesn't fit - do what you do when it doesn't} else if(f <
g)then !slow comparison, but you rarely get to here {do what
you do when it fits} else {do what you do when it doesn't} (of
course, if the original test was the other way around, you make
the appropriate changes in the above).So fast comparisons are
made almost always, but the exact comparisonis there precisely
when the fast comparisons are inadequate.Instead of running in
say 28% of the time, this might take more like42% of the time,
but it will always fit something when it should fitand never
fit something it shouldn't.Glen===
===
Subject:
: Open setIn
R^n,Complement of Open set = Closed setComplement of Closed
set = Open setIs it correct ?If then, Can we say above is hold
for any topology space ?===
===
Subject:
: Re: Open set> In R^n,>
Complement of Open set = Closed set> Complement of Closed set
= Open set> Is it correct ?> If then, Can we say above is hold
for any topology space ?The definition of Closed set is that
its complement is open. So thatis true by definition in any
topological space.A topological space is a set X together with
a collection T of subsetsof X satisfying:the null set an X are
in Tthe union of any collection of sets in T is also in Tthe
intersection of any pair of sets in T is also in TT is the
collection of open sets in the topological space.In Topology,
open sets do not have to be sets which areneighborhoods of
each of their points. You can have, for example, thesimplest
topology with only the null set and X being open, or you
canhave the densest topology with every subset of X being
open. Then nosets are closed.-Greg===
===
Subject:
: Re: Open set>
the densest topology with every subset of X being open. Then
no> sets are closed.No, all sets are closed.-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re: Open
set> In R^n,> Complement of Open set = Closed set> Complement
of Closed set = Open set> Is it correct ?> If then, Can we say
above is hold for any topology space ?Erm, yeah, that's true by
definition in any topological space :-)-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92;
mistakes his penis for a square root, 88-9Francis Wheen, _How
Mumbo-Jumbo Conquered the World_===
===
Subject:
: Re: Open setmy
several variable calculus classes have just started, and I
washoping if you could clarify something a little. you see, we
have juststarted, so for us, the definition of a neighborhood
of a point a issimple enough, |x-a| < r, for some real r in
|R^n. the open set is aset where each point has a neighborhood
contained in the set. a closedset is the complement of an open
set, which I suppose will not changeover time.but my (bad, I
know by now) habit of reading ahead is what getting meconfused
at this point. I know in |R^n any metric will do to define
anbd, so I am not worried about it at all. but the transition
towards ahigher level of abstraction is anticipated, where the
concept of openset would outperform that of nbd, right?what I
gather from miscellinous discrete readings is this:a
topological space is defined as a collection of open sets, the
termopen being sort of undefined, so that it follows the rules
that anyintersection and (infinite) union of open sets is
still open.how is this definition going to make sense? along
with the definitionof a continuous function as a function
whose inverse image of an openset is open? could you give me
some clue as to how to get high atthis?to clarify my position
here, I am confused about the ordering of thereal numbers that
looks so unavoidable in the epsilon-deltadefinition, which I
understand is equivalent to the description interms of open
sets as defined in my textbook, that is, along withnbd's that
make use of the ordering. the concept of choosing a verysmall
interval here to get a very small interval there does not
seemto make sense in the abstract definition I guess.so the
question boils down to: what exactly is retained and what
isstripped out in the transition?any help is deeply
appreciated.===
===
Subject:
: Re: Open set> what I gather from
miscellinous discrete readings is this:> a topological space
is defined as a collection of open sets, the term> open being
sort of undefined, so that it follows the rules that any>
intersection and (infinite) union of open sets is still open.>
how is this definition going to make sense? along with the
definition> of a continuous function as a function whose
inverse image of an open> set is open? could you give me some
clue as to how to get high at> this?The open sets of R^n have
the properties: any finite intersection andany infinite union
of open sets is open. So it is an example of atopology in the
more abstract definition.> to clarify my position here, I am
confused about the ordering of the> real numbers that looks so
unavoidable in the epsilon-delta> definition, which I
understand is equivalent to the description in> terms of open
sets as defined in my textbook, that is, along with> nbd's
that make use of the ordering. the concept of choosing a very>
small interval here to get a very small interval there does not
seem> to make sense in the abstract definition I guess.So you
will get to a theorem that says: a function f : R^n -> R^mis
continuous if and only if the inverse image of an open setis
open. So continuous functions defined in the epsilon-deltaway
are a special case of those defined in the abstract
way.Supersedes:
<physics-faq/criticism/galilean-invariance_1078308364@
rtfm.mit.eduX-Last-Updated: 1999/08/06===
===
Subject:
: Invariant
Galilean Transformations (FAQ) On All LawsSummary: All
laws/equations are Galilean invariant when expressed in the
generalized cartesian coordinates demanded by basic analytic
geometry, vector algebra, and measurement theory.Originator:
faqserv@penguin-lust.MIT.EDUDisclaimer: approval for *.answers
is based on form, not content. Opponents of the content should
first actually find out what it is, then think, then
request/submit-to arbitration by the appropriate neutral
mathematics authorities. Flaming the hard- working, selfless,
*.answers moderators evidences ignorance and despicable
netiquette.Archive-Name:
physics-faq/criticism/galilean-invarianceVersion:
0.04.03Posting-frequency: 15 days Invariant Galilean
Transformations (FAQ) On All Laws (c) Eleaticus/Oren C.
Webster Thnktank@concentric.netAn obvious typo or two
corrected.The Brittanica section revised to
less'pussy-footing' and to more directlyanticipate the
elementary measurementtheory and basic analytic geometrythat
is applied to the
transformationconcept.------------------------------===
===
Subject:

: 1. PurposeThe purpose of this document is to provide the
student of Physics,especially Relativity and Electromagnetism,
the most basic princ-iples and logic with which to evaluate the
historic justificationof Relativity Theory as a necessary
alternative to the classicalphysics of Newton and Galileo.We
will prove that all laws are invariant under the
Galileantransformation, rather than some being non-invariant,
afterwe show you what that means.We shall also show that
another primal requirement that SRexist is nonsense:
Michelson-Morley and Kennedy-Thorndike doindeed fit Galilean
(c+v) physics.------------------------------===
===
Subject:
: 2.
Table of Contents 1. Foreword and Intent 2. Table of Contents
3. The Principle of Relativity 4. The Encyclopedia Brittanica
Incompetency. 5. Transformations on Generalized Coordinate
Laws 6. The data scale degradation absurdity. 7. The
Crackpots' Version of the Transforms. 8. What does sci.math
have to say about x0'=x0-vt? 9. But Doesn't x.c'=x.c? 10. But
Isn't (x'-x.c')=(x-x.c) Actually Two Transformations? 11. But
Doesn't (x'-x.c+vt) Prove The Transformation Time Dependent?
12. But Isn't (x'-x.c')=(x-x.c) a Tautology? 13. But Isn't
(x'-x.c')=(x-x.c) Almost the Definition of a Linear Transform?
14. But The Transform Won't Work On Time Dependent Equations?
15. But The Transform Won't Work On Wave Equations? 16. But
Maxwell's Equations Aren't Galilean Invariant? 17. First and
Second Derivative differential
equations.------------------------------===
===
Subject:
: 3. The
Principle of Relativity and TransformationIf a law is
different over there than it is here,it is not one law, but at
least two, and leaves usin doubt about any third location. This
is thePrinciple of Relativity: a natural law must be thesame
relative to any location at which a given eventmay be
perceived or measured, and whether or not theobserver is
moving.The idea of location translates to a coordinatesystem,
largely because any object in motion couldbe considered as
having a coordinate system originmoving with it. If you
perceive me moving relativeto you - who have your own
coordinate system - willyour measurements of my position and
velocity fitthe same laws my own, different measurements
fit?If a law has the same form in both cases it iscalled
covariant. If it is identical in form, var-ables, and output
values, it is called invariant.What we're asking is that if
the x-coordinate, x,on one coordinate axis works in an
equation, doesthe coordinate, x', on some other, parallel
axiswork? Speaking in terms of the axis on which x isthe
coordinate, x' is the 'transformed' coordinate.The situation
is complicated because we're talkingabout coordinates -
locations - but in most mean-ingful laws/equations, it is
lengths/distances (andtime intervals) the equations are about,
and x coord-inates that represent good, ratio scale measures
ofdistances are only interval scale measures on the x'axis.
[See Table of Contents for discussion of scales.]So, if we
have an x-coordinate in one system, thenwe can call the x'
value that corresponds to the samepoint/location the transform
of x.In particular, the Principle of Relativity is embodiedin
the form of the Galilean transformation, whichrelates the
original x, y, z, t to x', y', z', t' bythe transform
equations x'=x-vt, y'=y, z'=z, t'=t inthe simplified case
where attention is focused onlyon transforming the x-axis, and
not y and z. In thecase of Special Relativity, the x' transform
is thesame except that x' is then divided by
sqrt(1-(v/c)^2),and t'=(t-xv/cc)/sqrt(1-(v/c)^2). In either
case, vis the relative velocity of the coordinate systems;if
there is already a v in the equations being trans-formed use u
or some other variable
name.------------------------------===
===
Subject:
: 4. The
Encyclopedia Brittanica Incompetency.One example of the
traditional fallacious ideathat an equation is not invariant
under the galileantransformation comes from the Encyclopedia
Brittanica:Before Einstein's special theory of relativitywas
published in 1905, it was usually assumedthat the time
coordinates measured in all inertialframes were identical and
equal to an 'absolutetime'. Thus, t = t'. (97)The position
coordinates x and x' were thenassumed to be related by x' = x
- vt. (98)The two formulas (97) and (98) are called aGalilean
transformation. The laws of nonrelativ-istic mechanics take
the same form in all framesrelated by Galilean
transformations. This is therestricted, or Galilean, principle
of relativity.The position of a light wave front speeding
fromthe origin at time zero should satisfy x^2 - (ct)^2 = 0
(99)in the frame (t,x) and (x')^2 - (ct')^2 = 0 (100)in the
frame (t',x'). Formula (100) does nottransform into formula
(99) using the transform-ations (97) and (98),
however..................................................
Besides the trivially correct statement of what theGalilean
'transform' equations are, there is exactlyone thing they got
right.I. Eq-100 is indeed the correct basis for discussing the
question of invariance, given that eq-99 is the correct
'stationary' (observer S) equation. [Let observer M be the
'moving'system observer.] In particular, eq-100 is of exactly
the same form [the square of argument one minus the square of
argument two equals zero (argument three).]II. It is nonsense
to say eq-99 should be derivable from eq-100; for one thing,
the transforms are TO x' and t' from x and t, not the other
way around, and the idea that either observer's equation
should contain within itself the terms to simplify or
rearrange to get to the other is ridiculous. As the transform
equations say, the relationship of t', x' to t, x is based on
the relative velocity between the two systems, but neither the
original (eq-99) equation nor the M observer equation is about
a relationship between coordinate systems or observers. One
might as well expect the two equations to contain banana
export/import data; there is no relevancy. The 'transform'
equations are the relationships between x' and x, t' and t and
have nothing to do with what one equation or the other ought to
'say'. The equations' content is the rate at which light
emitted along the x-axes moves.III. Most remarkable, the True
Believer SR crackpots who most despise the consequences of
measurement theory (demonstrable fact) contained in this
document are those who want to argue against our saying the
Britt- anica got eq-100 right; They insist that the correct
equation is derived directly from x'=x-vt and t'=t. Solve for
x=x'+vt and replace t with t', then substitute the result in
eq-99: (x'+vt')^2 - (ct')^2 = 0. Besides the fact that this
results in an equation with arguments exactly equal to eq-99,
they will insist the transform is not invariant.IV. A major
justification they have for their idea of the correct M system
equation on which to base the the discussion of invariance, is
that the variables are M system variables, never mind the fact
that the arguments are S system values. That argument of theirs
is arrant nonsense. The velocity v that S sees for the M system
relative to herself is the negative of what the M system sees
for the S system relative to himself. In other words, x'+vt'
is a mixed frame expression and it is x'+(-v)t' that would be
strictly M frame notation, and that equation is far off base.
[Work it out for yourself, but make sure you try out an S
frame negative v so as not to mislead yourself.]V. In I. we
said: given that eq-99 is the correct 'stationary' equation.
Let's look at it closely: x^2 - (ct)^2 = 0 (99) This whole
matter is supposed to be about coordinate transforms. Is that
what t is, just a coordinate? No. It isn't, in general.
Suppose you and I are both modelling the same light event and
you are using EST and I'm using PST. 'Just a time coordinate'
is just a clock reading amd your t clock reading says the
light has been moving three hours longer than my clock reading
says. Well, that's what the idea that t is a coordinate means.
Eq-99 works if and only if t is a time interval, and in
particular the elapsed time since the light was emitted. Thus,
that equation works only if we understand just what t is, an
elapsed time, with emissioon at t=0. However, we don't have to
'understand' anything if we use a more intelligent and
insightful form of the equation: (x)^2 - [ c(t-t.e) ]^2 = 0,
where t.e is anyone's clock reading at the time of light
emission, and t is any subsequent time on the same clock.
Similarly, x is not just a coordinate, but a distance since
emission. (x-x.e)^2 - [ c(t-t.e) ]^2 = 0 (99a)VI. In the
spirit of 'there is exactly one thing they got right', the
correct M system version of eq-99a is eq-100a: (x'-x.e')^2 - [
c(t'-t.e') ]^2 = 0 (100a) Every observer in the universe can
derive their eq-100a from eq-99a and vice versa, not to
mention to and from every other observer's eq-99a. Now, THAT's
invariance. [You do realize that every eq-100a reduces to
eq-99a, when you back substitute from the transforms, right?
t.e'=t.e,
x.e'=x.e-vt.]------------------------------===
===
Subject:
: 5.
Transformations on Generalized Coordinate LawsThe traditional
Gallilean transform is correct: t' = t x' = x - vt.But
remember this: a transform of x doesn't effectjust some values
of x, but all of them, whether theyare in the formula or not.
This is important if youwant to do things right. The crackpot
position isstrongly against this sci.math verified position,
andthe apparently standard coordinate
pseudo-transformationthey suggest is perhaps the result. {See
Table ofContents.]Let's use a simple equation: x^2 + y^2 =
r^2, which isthe formula for a circle with radius r, centered
at alocation where x=0.But what if the circle center isn't at
x=0? Well, we'dwant to use the form analytic geometry, vector
algebra,and elementary measurement theory tells us to use, a
formwhere we make explicit just where the circle center
is,even if it is at x=x0=0: (x-x0)^2 + (y-y0)^2 = r^2.The
circle center coordinate, x0, is an x-axis coordinate,just
like all the x-values of points on the circle.So, in proper
generalized cartesian coordinate formsof laws/equations we
want to transform every occurenceof x and x0 - by whatever
name we call it: x.c, x_e,whatever.So, what is the transformed
version of (x-x0)? Why,(x'-x0'); both x and x0 are
x-coordinates, and everyx-coordinate has a new value on the
new axis.So, what is the value of (x'-x0') in terms of the
originalx data?is also true for x0'=x0-vt: (x'-x0')=[
(x-vt)-(x0-vt) ]=(x-x0).In other words, when we use the
generalized coordinate formspecified by analytic geometry, we
find that the value of(x'-x0') does not depend on either time
or velocity in anyway, shape, form, or fashion.Similarly for
(y-y0).We can treat time the same way if necessary: (t-t0).The
above is a proof that any equation in x,y,z,t isinvariant under
the galilean transforms. Just use thegeneralized coordinate
form, with (x-x0)/etc, in thetransformation process, not the
incompetently selectedprivileged form, with just x/etc.[The
form is privileged because it assumes the circlecenter, point
of emission, whatever, is at the origin ofthe axes instead at
some less convenient point. Aftertransform the coordinate(s)
of the circle center/originare also changed but the privileged
form doesn't makethis explicit and screws up the calculations,
whichshould be based on (x'-x0') but are calculated as
(x'-0).]The value of (x'-x0') is the same as (x-x0). That
makessense.Draw a circle on a piece of paper, maybe to the
rightside of the paper. On a transparent sheet, draw x and
ycoordinate axes, plus x to the right, plus y at the top.Place
this axis sheet so the y-axis is at the left sideof the circle
sheet.Now answer two questions after noting the x-coordinate
ofthe circle center and then moving the axis sheet to the
right:(a) did the circle change in any way because you
movedthe axis sheet (ie because you transformed the
coordin-nate axis)?(b) did the coordinate of the circle center
change?The circle didn't change [although SR will say it
did];that means that (x'-x0') does indeed equal (x-x0).The
coordinate of the circle center did change, and itchanged at
the same rate (-vt) as did every point onthe circle. That
means that x0'<>x0, and the fact thecircle center didn't
change wrt the circle, means thatthe relationship of x0' with
x0 is the same as that ofany x' on the circle with the
corresponding x: x'=x-vt;x0'=x0-vt.This is to prepare you for
the True Believer crackpots thatsay 'constant' coordinates
can't be transformed; some evensay they aren't coordinates.
These crackpots include somethat brag about how they were
childhood geniuses, btw.QED: The galilean transformation for
any law ongeneralized Cartesian coordinates is invariant
underthe Galilean transform.The use of the privileged form
explains HOW the transformedequation can be messed up, the
next 
===
Subject:
 explains whatthe screwed up effect of the
transform is, and how useof the generalized form corrects the
screwup.------------------------------===
===
Subject:
: 6. The data
scale degradation absurdity.The SR transforms and the Galilean
transforms bothconvert good, ratio scale data to inferior
intervalscale data. The effect is corrected, allowed for,when
the transforms are conducted on the generalizedcoordinate
forms specified by analytic geometry andvector algebra.Both
sets of transforms are 'translations' - lateralmovements of an
axis, increasing over time in thesecases - but with the SR
transform also involving arescaling. It is the translation
term, -vt in the xtransform to x', and -xv/cc in the t
transform to t',that degrades the ratio scale data to interval
scaledata. In general, rescaling does not effect scalequality
in the size-of-units sense we have here.SR likes to consider
its transforms just rotations,however - in spite of the fact
Einstein correctly saidthey were 'translations' (movements) -
and in the caseof 'good' rotations, ratio scale data quality
is indeedpreserved, but SR violates the conditions of good
ro-tations; they are not rigid rotations and they
don'tappropriately rescale all the axes that must be
rescaledto preserve compatibility.The proof is in the pudding,
and the pudding is thecombination of simple tests of the
transformations.We can tell if the transformed data are ratio
scaleor interval.Ratio scale data are like absolute Kelvin. A
measure-ment of zero means there is zero quantity of thestuff
being measured. Ratio scale data support add-ition,
subtraction, multiplication, and division.The test of a ratio
scale is that if one measurelooks like twice as much as
another, the stuffbeing measured is actually twice as much.
Withabsolute Kelvin, 100 degrees really is twice theheat as 50
degrees. 200 degrees really is twiceas much as 100.Interval
scale data are like relative Celsius, whichis why your science
teacher wouldn't let you use itin gas law problems. There is
only one mathematicaloperation interval scales support, and
that has tobe between two measures on the same scale:
subtraction.100 degrees relative (household) Celsius is not
twiceas much as 50; we have to convert the data to
absoluteKelvin to tell us what the real ratio of
temperaturesis.However, whether we use absolute Kelvin or
relativeCelsius, the difference in the two temperature
readingsis the same: 50 degrees.Thus, if we know the real
quantities of the 'stuff'being measured, we can tell if two
measures are ona ratio scale by seeing if the ratio of the
twomeasures is the same as the ratio of the known
quant-ities.If a scale passes the ratio test, the interval
scale testis automatically a pass.If the scale fails the ratio
test, the interval scaletest becomes the next in line.It isn't
just the bare differences on an intervalscale that provides
the test, however. Differencesin two interval scale measures
are ratio scale, soit is ratios of two differences that tell
the tale.Let's do some testing, and remember as we do that
ourconcern is for whether or not the data are messed up,not
with 'reasons', excuses, or
avoice.------------------------------------------------------
Are we going to take a transformed length (difference)and see
whether that length fits ratio or interval scaledefinitions?Of
course, not. Interval scale data are ratio afterone measure is
subtracted from another. That is themajor reason the SR
transforms can be used in science.Let there be three rods, A,
B, C, of length 10, 20, 40,respectively. These lengths are on
a known ratio scale,our original x-axis, with one end of each
rod at theorigin, where x=0, and the other end at the
coordinatethat tells us the correct lengths.Note that these
x-values are ratio scale only becauseone end of each rod is at
x=0. That may remind you ofthe correct way to use a ruler or
yard/meter-stick:put the zero end at one end of the thing you
aremeasuring. Put the 1.00 mark there instead of the zero,and
you have interval scale measures.Let A,B,C, be 10, 20, 40.Let
a,b,c be x' at v=.5, t=10.x'=x-vt.A B C a b c----------------
--------------------10 20 40 5 15 35----------------
--------------------B/A = 2 b/a = 3C/A = 4 c/a = 7C/B = 2 c/b
= 2.333 Obviously, the transformed values are no longer ratio
scale. The effect is less on the greater values.C-A = 10 b-a =
10C-A = 30 c-a = 30C-B = 20 c-b = 20 Obviously, the transformed
values are now interval scale. This will hold true for any
value of time or velocity.(C-A)/(B-A) = 3 (c-a)/(b-a) =
3(C-B)/(B-A) = 2 (c-b)/(b-a) = 2 Obviously, the ratios of the
differences are ratio scale, being identical to the ratios of
the corresponding original - ratio scale - differences.The
main difference between these results and the SRresults is
that the differences do not correspond soneatly to the
original, ratio scale, differences.This is due only to the
rescaling by 1/sqrt(1-(v/c)^2).The ratios of the differences
on the transformed valuesdo correspond neatly and exactly to
the ratio scaleresults.Using the generalized coordinate form,
such as (x-x0),the transform produces an interval scale x' and
aninterval scale x0'. That gives us a ratio scale (x'-x0'),just
like - and equal to -
(x-x0).------------------------------===
===
Subject:
: 7. The
Crackpots' Version of the Transforms.It has become apparent -
whether misleading or not -that the crackpot responses to the
obvious derive froma common source, whether it be bandwagoning
or theirSR instructors.Below, in the sci.math subject, we see
that all sci.mathrespondents agree with the basic
controversial positionof this faq: every coordinate is
transformed, whether asupposed constant or not.Think about it,
the generalized coordinate of a circlecenter, x0, applies to
infinities upon infinities ofcircle locations (given y and z,
too); it is a constantonly for a given circle, and even then
only on a givencoordinate axis.And even variables are often
held 'constant' duringeither integration or
differentiation.The utility of a variable is that you can
discuss allpossible particular values without having to single
outjust one. That utility does not make particular - singledout
- values on the variable's axis not values of thevariable just
because they have become named values.In any case, all that is
preamble to the incompetent ideathey have proposed for a
transform of coordinates. It isbased on the idea that the
circle center, point of emission,whatever, has coordinates
that cannot be transformed.Let there be an equation, say (x)^2
- (ict)^2 = 0.What is the transformed version of that
equation?Answer: (x')^2 - (ict')^2 = 0. That's the one thing
theBrittanica got right. Note that the leading crackpot
justcriticized this faq for presuming to correct the
Britt-anica, but it then and before poses the incompetent
pseudo-transform we analyze here in this section.x to x' and t
to t' are obviously coordinate transforms;the x and t
coordinates have been replaced by the coord-inates in the
primed system.A tranform of an equation from one coordinate
system toanother is NOT a substitution of the/a definition of
xfor itself; that is not a coordinate transformation.The most
that can said for such a substitution is thatit is a change of
variable.But the crackpots are calling this a coordinate
trans-form of the original equation: (x'+vt)^2 - (ict')^2 =
0.It is not a coordinate transform, of course,
exceptaccidentally. (x'+vt) is not the primed
systemcoordinate, it is another form/expression of x. Theyget
that substitution by solving x'=x-vt for x; x=x'+vt.So, by
incompetent misnomer, they accomplish what theyhave been
railing against all along.It has been the generalized
coordinate form in question allthis time: (x-x0)^2 - (ict)^2 =
0.Here they substitute for x instead of transforming to
theprimed frame: (x'+vt-x0)^2 - (ict')^2. ----- ^ | ^ |It is
still x ^ but see what they have accomplishedby their
mis/malfeasance: [x'+vt-x0]=[x'+(vt-x0)]=[x'-(x0-vt)].
=[x'-x0']The crackpots have been bragging about how you
don'thave to transform the circle center's coordinate
totransform the circle center's coordinate. Braggingthat what
they were doing was not what they saidthey were doing.This
does give us insight as to some of the crackpotvariations on
their x0'<>x0-vt theme, which in all thevariations will be
discussed in later sections..They are used to seeing the mixed
coordinate form,(x'+vt-x0) without realizing what it
respresented,so - accompanied with a lack of understanding
ofthe term 'dependent' - they are used to seeing justthe one
vt term, and not the one hidden in the defi-nition of x' and
are used to imagining it makes thewhole expression time
dependent and thus not invariant.About which, let x=10, let,
x0=20, v=10, and tvariously 10 and 23:(x-x0)=-10. Using their
(x'+vt-x0):For t=10, we have (x'+vt-x0) = [ (10-10*10) +
(10*10) - (20) ] = -90 + 100 - 20 = -10 = (x-x0)For t=23, we
have (x'+vt-x0) = [ (10-10*23) + (10*23) - (20) ] = -220 + 230
- 20 = -10 = (x-x0)The result depends in no way on the value of
time;we showed the obvious for a couple of instances of tjust
so you can see that the crackpots not only donot understand
the obvious logic of the algebra{ (x'-x0')=[ (-vt)-(x0-vt)
]=(x-x0) } - which showsthat the transform has no possible
time term effect -but they don't understand even a simple
arithmeticdemonstration of the facts.Oh. Their (x'+vt-x0) or
(x'+vt'-x0) reduces the sameway since t'=t:
(x-vt+vt-x0)=(x-x0).Their process, which says (x'+vt') is the
transformof x, says that (x'+vt') is the moving system
locationof x, but it can't be because x is moving further
inthe negative direction from the moving viewpoint.That
formula will only work out with v<0 which is indeedthe
velocity the primed system sees the other moving at.However,
that formula cannot be derived from x'=x-vt,the formula for
transformation of the coordinates fromthe unprimed to the
primed,------------------------------===
===
Subject:
: 8. What does
sci.math have to say about x0'=x0-vt?The crackpots'
positions/arguments were put to sci.mathin such a way that at
least two or three who posted re-sponses thought it was your
faq-er who was on the idiot'sside of the questions.Their
responses:----------------------------------------------------
------I. x0' = x0. In other words: x0' <> x0-vt, or constant
values on the x-axis are not subject to the transform.AA:
==============================================================
====== No. x0' = x0 - vt. Well, if you want, you could define
constant values on the x-axis, butin the context of the
question that is not relevant. The relevant fact isthat if the
unprimed observer holds an object at point x0, then theprimed
observer assigns to that object a coordinate x0' which
isnumerically related to x0 by x0'= x0 -vt.AA:
==============================================================
======EE:
==============================================================
======What does this mean? The line x=x0 will give
x'=x-v*t=x0-vt', so if x0'is to give the coordinate in the
(x',t',)-system, it will be given byx0'=x0-v*t': ie., it is
not given by a constant. Thus, being at rest(constant
x-coordinate) is a coordinate-dependent concept.EE:
==============================================================
======GG:
==============================================================
======Sounds very false. We can say that the representation of
the point X0 isthe number x0 in the unprimed system, and x0' in
the primed system.Clearly x0 and x0' are different, if vt is
not zero. However one may saythat (though it sounds/is stupid)
the point X0 itself is the samethroughout the transformation.
However that expression soundsmeaningless, since a transform
(ok, maybe we should call it a change ofbasis) is only a
function that takes the point's representation in onesystem
into the same point's representation in another system. It
ispreferrable to use three notations: X0 for the point itself
and x0 andx0' for the points' representations in some
coordinate systems.GG:
==============================================================
======------------------------------===
===
Subject:
: 9. But Doesn't
x.c'=x.c?That idea is one of the most idiotic to come up, and
it doesso frequently. And in a number of guises.The idea being
that x.c' <> x.c-vt, with x.c being whatwe have called x0
above; the notation makes no difference.Some crackpots have
managed to maintain that position evenafter graphs have
illustrated that such an idea means thatafter a while a circle
center represented by x.c' could beoutside the circle.The
leading crackpot just make that explicit, as far asone can
tell from his befuddled post in response to a lineabout active
transforms, which are actually moving bodysituations, not
coordinate
transformations:----------------------------------------------
----------------------e>An active transform is not a
coordinate transform, ... Right, it is a transform of the
center (in the opposite direction) done to effect the change
of coordinates without a coordinate transform. ...E: Transform
of the center? Center of a circle? He really is saying a circle
center moves in the opposite direction of the circle!
Right?--------------------------------------------------------
------------If r=10 and x.c was at x.c=0, then the points on
the circle(10,0), (-10,0), (0,10) and (0,-10) could at some
time become(-10,0), (-30,0), (-20,10), and (-20,-10), but with
x.c'=x.c,the circle center would be at (0,0) still! The circle
is herebut its center is way, way over there! Indeed, although
a changeof coordinate systems is not movement of any object
described inthe coordinates, the x.c'=x.c crackpottery is
tantamount to thecircle staying put but the center moving
away. Or vice versa.------------------------------===
===
Subject:
:
10. But Isn't (x'-x.c')=(x-x.c) Actually Two
Transformations?One crackpot puts the (x'-x.c')=(x-vt -
x.c+vt) relationshiplike this: (x-vt+vt - x.c).See, he says,
that is transforming x (with x-vt - x.c) and thenreversing the
transform (x-vt+vt - x.c).That's just another crackpot form of
the idiocy thatx.c' <> x.c-vt. You'll have noticed the
implicationis that there is no transform vt term relating to
x.c.------------------------------===
===
Subject:
: 11. But Doesn't
(x'-x.c+vt) Prove The Transformation Time Dependent?That
particular crackpottery is perhaps more corrupt thanmoronic,
since it includes deliberately hiding a vt term fromview, and
pretending it isn't there. [However, we have seenabove that it
is a familiar incompetency, and not likely anoriginal.]Look,
the crackpots say, there is a time term in thetransformed (x'
- x.c+vt). The transform isn't invariant!It's time
dependent!Just put x' in its original axis form, also, which
revealsthe other time term, the one they hide: (x'-x.c+vt) =
(x-vt - x.c+vt) = (x-x.c).So, at any and all times, the
transform reduces to theoriginal expression, with no time term
on which to bedependent.Then there is the fact that if you
leave the equationin any of the various notation forms - with
or withoutreducing them algebraicly - the arithmetic always
comesdown to the same as (x-x.c). That means nothing to
crack-pots, but may mean something to
you.------------------------------===
===
Subject:
: 12. But Isn't
(x'-x.c')=(x-x.c) a Tautology?My dictionary relates
'tautology' to needless repetition.That's another form of the
x.c' <> x.c-vt idiocy.The repetition involved is the vt
transformation term.Apply the -vt term to the x term, and it
is needlessrepetition to apply it anywhere again? The 'again'
isto the x.c term. The x.c' = x.c crackpot idiocy.The
repetition of the vt terms is required by the presenceof two x
values to be transformed.Be sure to note the next
section.------------------------------===
===
Subject:
: 13. But
Isn't (x'-x.c')=(x-x.c) Almost the Definition of a Linear
Transform?Now, how on earth can we relate a tautology to a
basicdefinition in math?we get this
definition:---------------------------------------------------
-----------A linear transformation, A, on the space is a
method of corr-esponding to each vector of the space another
vector of thespace such that for any vectors U and V, and any
scalarsa and b, A(aU+bV) = aAU +
bAV.----------------------------------------------------------
---Let points on the sphere satisfy the vector X={x,y,z,1},and
the circle center satisfy C={x.c,y.c,z.c,1}. Let a=1,and
b=-1.Let A= ( 1 0 0 -ut ) ( 0 1 0 -vt ) ( 0 0 1 -wt ) ( 0 0 0
1 )A(aX+bC) = aAX + bAC. aX+bC = (x-x.c, y-y.c, z-z.c, 0 ).The
left hand side: A( x - x.c , y - y.c, z - z.c, 0 ) = ( x-x.c ,
y-y.c, z-z.c, 0 ).The right hand side: aAX= ( x-ut, y-vt,
z-wt, 1 ). bAC= (-x.c+ut, -y.c+vt, -z.c+wt, -1 ).and aAX+bAC =
( x-x.c, y-y.c, z-z.c, 0 ).Need it be said?Sure: QED. On the
galilean transform thedefinition of a linear transform,
A(aU+bV)=aAU + bAV,is completely satisfied.The generalized
form transforms exactly andnon-reduntly - with ONE TRANSFORM,
not atransform and reverse transform - and non-tautologically,
just as the very definitionof a linear transform says it
should.And does so with absolute invariance, with thisgalilean
transformation.------------------------------===
===
Subject:
: 14.
But The Transform Won't Work On Time Dependent Equations?The
main crackpot that has asserted such a thing was referringto
equations such as in 
===
Subject:
 4, above. The Light
Sphereequation; for which we have shown repeatedly elsewhere
that thenumerical calculations are identical for any primed
values asfor the unprimed values.The presence - before
transformation - of a velocity termseems to confuse the
crackpots. It turns out there is ex-treme historical reason
for this, as you will see in thesubject on Maxwell's
equations.------------------------------===
===
Subject:
: 15. But
The Transform Won't Work On Wave Equations?See 
===
Subject:
 17,
below, for a discussion of Second Derivativeforms and the
galilean transforms.------------------------------===
===
Subject:
:
16. But Maxwell's Equations Aren't Galilean Invariant?Oh? Just
what is the magical term in them that prevents(x'-x.c')=(x-vt -
x.c+vt)=(x-x.c) from holding true?It turns out not to be magic,
but reality, that interfereswith the application of the
galilean transforms to the gen-eralized coordinate form(s) of
Maxwell: there are no coordi-nates to transform!When True
Believer crackpots are shown the simpledemonstration that the
galilean transform ongeneralized cartesian coordinates is
invariant,their first defense is usually an incredibly
stupidx0'=x0, because the coordinate of a circle center,or
point of emission, etc, is a constant and can'tbe
transformed.The last defense is but Maxwell's equations are
notinvariant under that coordinate transform. Whenasked just
what magic occurs in Maxwell that wouldprevent the simple
algebra (x'-x0')=[ (x-vt)-(x0-vt) ]=(x-x0)from working, and
when asked them for a demonstration,they will never do so,
however many hundreds oftimes their defense is asserted.The
reason may help you understand part of Einstein's1905 paper in
which he gave us his absurd SpecialRelativity derivation:THERE
ARE NO COORDINATES IN THE EQUATIONS TO BE TRANSFORMED.Einstein
gave the electric force vector as E=(X,Y,Z)and the magnetic
force vector as B=(L,M,N), where theforce components in the
direction of the x axis areX and L, Y and M are in the y
direction, Z and N inthe z direction.Those values are not,
however, coordinates, but valuesvery much like acceleration
values.BTW, the current fad is that E and B are 'fields',
havingbeen 'force fields' for a while, after being
'forces'.So, when Einstein says he is applying his
coordinatetransforms to the Maxwell form he presented, he
iseither delusive or lying.(a) there are no coordinates in the
transform equations he gives us for the Maxwell transforms,
where B=beta=1/sqrt(1-(v/c)^2): X'=X. L'=L. Y'=B(Y-(v/c)N).
M'=B(M+(v/c)Z). Z'=B(Z+(v/c)M). N'=B(N-(v/c)Y). X is in the
same direction as x, but is not a coordinate. Ditto for L.
They are not locations, coordinates on the x-axis, but force
magnitudes in that direction. Similarly for Y and M and y, Z
and N and z.(b) the v of the coordinate transforms is in
Maxwell before any transform is imposed; Einstein's transform
v is the velocity of a coordinate axis, not the velocity he
touched it.(c) if they were honest Einsteinian transforms,
they'd be x, which means it is X and L that are supposed to be
transformed, not Y and M, and Z and N. And when SR does
transform more than one axis, each axis has its own velocity
term; using the v along the x-axis as the v for a y-axis and
z-axis transform is thus trebly absurd: the axes perpendicular
to the motion are not changed according to SR, the v used is
not their v, and the v is not a transform velocity anyway.(d)
as everyone knows, the effect of E and B are on the direction.
Both the speed and direction are changed by E and B, but v -
the speed - is a constant in SR.As absurd as are the
previously demonstrated Einsteinianblunders, this one
transcends error and is an incredibleexample of True Believer
delusion propagating over decades.The components of E and B do
differ from point to point,and in the variations that are not
coordinate free,they are subject to the usual invariant
galilean trans-formation when put in the generalized
coordinate
form.---------------------------------------------------------
----The SR crackpots don't know what coordinates are.
Thevarious things they call coordinates include coordin-nates,
but also include a variety of other
quantities.---------------------------------------------------
---1. One may express coordinates in a one-axis-at-a-time
manner [like x^2+y^2=r^2] but it is the use of vector notation
that shows us what is going on. In vector notation the triplet
x,y,z [or x1,x2,x3, whatever] represents the three spatial
coordinates, but there are so-called basis vectors that
underlie them. Those may be called i,j,k. Thus, what we
normally treat as x,y,z is a set of three numbers TIMES a
basis vector each.2. These e*i, f*j, g*k products can have a
lot of meanings. If e, f, j are distances from the origin of
i,j,k then e*i, f*j, g*k are coordinates: distances in the
directions of i,j,k respectively, from their origin. That
makes the triplet a coordinate vector that we describe as
being an x,y,z triplet; perhaps X=(x,y,z). The e*i, f*j, g*k
products could be directions; take any of the other vectors
described above or below and divide the e,f,g numbers by the
length of the vector [sqrt(e^2+f^2+g^2)]. That gives us a
vector of length=1.0, the e,f,g values of which show us the
direction of the original vector. That makes the triplet a
direction vector that we describe as being an x,y,z triplet;
perhaps D=(x,y,z). The e*i, f*j, g*k products could be
velocities; take any of the unit direction vectors described
above and multiply by a given speed, perhaps v. That gives a
vector of length v in the direction specified. That makes the
triplet a velocity vector that we describe as being an x,y,z
triplet; perhaps V=(x,y,z). Each of the three values, e,f,g,
is the velocity in the direction of i,j,k respectively. The
e*i, f*j, g*k products could be accelerations; take any of the
unit direction vectors described above and multiply by a given
acceleration, perhaps a. That gives a vector of length a in
the direction specified. That makes the triplet an
acceleration vector that we describe as being an x,y,z
triplet; perhaps A=(x,y,z). Each of the three values, e,f,g,
is the acceleration in the direction of i,j,k respectively.
The e*i, f*j, g*k products could be forces (much like accel-
erations); take any of the unit direction vectors described
above and multiply by a given force, perhaps E or B. That
gives a vector of length E or B in the direction specified.
That makes the triplet a force vector that we describe as
being an x,y,z triplet; perhaps E=(x,y,z) or B=(x,y,z). Each
of the three values, e,f,g, is the force in the direction of
i,j,k respectively.Einstein's - and Maxwell's - E and B arenot
coordinate
vectors.======================================================
======There is another variety of intellectual befuddlement
thatmisinforms the idea that Maxwell isn't invariant under
thegalilean transform: confusions about velocities.Velocities
With Respect to Coordinate
Systems.-----------------------------------------------Aaron
Bergman supplied the background in a post to a
sci.physics.*newsgroup:=======================================
========================Imagine two wires next to each other
with a current I in each.Now, according to simple E&M, each
current generates a magneticfield and this causes either a
repulsion or attraction betweenthe wires due to the
interaction of the magnetic field and thecurrent. Let's just
use the case where the currents are parallel.Now, suppose you
are running at the speed of the current betweenthe wires. If
you simply use a galilean transform, each wire,having an equal
number of protons and electrons is neutral. So,in this frame,
there is no force between the wires. But this is
acontradiction.===============================================
=================First of all, the invariance of the galilean
transform (x'-x.c')=(x-x.c), insures that it is an error to
imagine there is anydifference between the data and law in one
frame and in another;the usual, convenient rest frame is the
best frame and only framerequired for universal analysis.
[Well, (x'<>x, x,c'<>x.c, but(x'-x.c')=(x-x.c).]Second, given
that you decide unnecessarily to adapt a law toa moving frame,
don't confuse coordinate systems with meaningfulphysical
objects, like the velocity relative to a coordinatesystem
instead of relative to a physical body or field.In other
words, what does current velocity with respect to acoordinate
system have to do with physics?Nothing. Certainly not anything
in the example Bergman gave.What is relevant is not current
velocity with respect to acoordinate system, but current
velocity with respect to wiresand/or a medium. The velocity of
an imaginary coordinate sys-tem has absolutely nothing to do
with meaningful physical vel-ocity. You can - if you are
insightful enough and don't violateitem (e) - identify a
coordinate system and a relevant physicalobject, but where
some v term in the pre-transformed law isin use, don't confuse
it with the velocity of the coordinatetransform.Velocities With
Respect to ...
What?-----------------------------------------------Albert
Einstein opened his 1905 paper on Special Relativitywith this
ancient
incompetency:=================================================
==============The equations of the day had a velocity term
that was takenas meaning that moving a magnet near a conductor
would createa current in the conductor, but moving a conductor
near awire would not. This was belied by fact, of course.The
important velocity quantity is the velocity of themagnet and
conductor with respect to each other, not tosome absolute
coordinate frame (as far as we know) andnot to an arbitrary
coordinate system.One possible cause was the idea: but the
equation says the magnetmust be moving wrt the coordinate
system or ... the absoluterest frame.There not being anything
in the equation(s) to say either ofthose, it is amazing that
folk will still insist the velocityterm has nothing to do with
velocity of the two bodies wrteach
other.--------------------------------------------------------
---------------------------------===
===
Subject:
: 17. First and
Second Derivative differential equations.One of the
intellectually corrupt ways ofdenying the very simple
demonstration ofgalilean invariance of all laws expressedin
the generalized coordinate form demandedby analytic geometry,
vector analysis, andmeasurement theory [ (x'-x.c')=[
(x-vt)-(x.c-vt) ]=(x-x.c) ]is the assertion that those
equations 'over there'(usually Maxwell or wave) are somehow
immune tothe elementary laws of algebra used to demon-strate
the invariance. [Unfortunately, theassertions are never
accompanied by referenceto the magical math that makes
elementary al-gebra invalid. Wonder why that is?]Part of the
time it is based on the old lorebased on the incompetent
transformation ofthe privileged form of an equation insteadof
the correct form. [Evidence of this isany reference to an
effect due to the velocityof the transform; it falls out
algebraicly- as you see above - and cancels out
arith-metically - as you can see above.]But usually it is just
whistling in the dark,waving the cross (zwastika, I'd say)
atthe mean old vampire.The most general equation that could be
conjuredup is a differential with either First or
SecondDerivatives.Let's examine the plausibility of such
magicalmagical, non-invariance assertions.(a) to get a Second
Derivative you must have a First Derivative.(b) to get a First
Derivative you must have a function to differentiate.(c) to get
a Second Derivative you must have a function in the second
degree.So, let us examine the question as to whetherany such
common Maxwell/wave equation willdiffer for(a) the common,
privileged form, represented as ax^2, with a being an unknown
constant function.(b) the generalized cartesian form,
represented as a(x-x.c)^2 = ax^2 -2ax(x.c) + ax.c^2, with a
being an unknown constant function.(c) the transformed
generalized cartesian form, represented as a(x-vt -x.c+vt)^2,
same as for (b), = ax^2 -2ax(x.c) + ax.c^2, of course, with a
being an unknown constant function.I. for (a), remembering
that x.c is a constant, and that this version is only correct
because x.c=0, otherwise (b) is the correct form: d/dx ax^2 =
2ax (d/dx)^2 ax^2 = 2aII. for (b), remembering that x.c is a
constant. d/dx (ax^2 -2ax(x.c) + ax.c^2) = 2ax - 2ax.c
(d/dx)^2 (ax^2 -2ax(x.c) + ax.c^2) = 2aIII. for (c); same as
for (b).So, what we have seen so far is(1) differential
equations in the second degree- the wave equations - must
clearly be the same forall forms: the privileged form in x,
the generalizedcartesian form in x and the centroid, x.c, or
thetransformed generalized cartesian form.That is, anyone who
imagines that correct usagegives different results for
galilean transformedframes is at first showing his ignorance,
and inthe end showing his intellectual corruption.(2) As far
as the First Derivatives are concerned, theonly cases in which
there really is a difference betweenthe two forms is where x.c
<> 0, and in that case, theuse of the privileged form is
obviously incompetent.So, how do you correctly use the
differential equations?If you are using rest frame data with
the centroidat x=0, etc, you can't go wrong without trying
togo wrong.If you are using rest frame data with the
centroidnot at x=0, you must use (x-x.c) anyplace x appearsin
the equation.If you are using moving frame data, you must use
themoving frame centroid as well as the light front(or
whatever) moving frame data itself, perhaps firstcalculating
(x'-x.c'), which equals (x-x.c) which isobviously correct, and
which is obviously the plain oldcorrect x of the privileged
form.Unless, of course, there really is some magical termor
expression that invalidates the obvious and elemen-tary
algebra of the invariance demonstration.Or maybe you just
whistle when you don't want basicalgebra to hold
true.Eleaticus!---?---!---?---!---?---!---?---!---?---!---?---
!---?---!---?---!---?! Eleaticus Oren C. Webster
ThnkTank@concentric.net ?! Anything and everything that
requires or encourages systematic ?! examination of premises,
logic, and conclusions
?!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
---!---?===
===
Subject:
: Re: Invariant Galilean Transformations
(FAQ) On All Laws> Invariant Galilean Transformations (FAQ) On
All Laws> (c) Eleaticus/Oren C. Webster>
Thnktank@concentric.net[snip 1300 lines of trolled
garbage]Originally trolled across sci.physics
sci.physics.relativityalt.physics sci.math sci.answers
alt.answers
news.answershttp://b5.sdvc.uwyo.edu/bab5/snds/
argcstpd.wavPsychotic ineducable boring troll Eleaticus, You
see yourself this
way,http://www.mazepath.com/uncleal/effete6.jpg The entire
remainder of the planet sees you this
way,http://www.mazepath.com/uncleal/effete3.pnghttp://
w0rli.home.att.net/youare.swfhttp://www.mazepath.com/uncleal/
sunshine.jpghttp://www.you-moron.com/http://www.apa.org/
journals/psp/psp7761121.htmlhttp://insti.physics.sunysb.edu/~
siegel/quack.html<http://www.firehead.org/~jessh/film/kubrick/
Kubrick-Psycho.html<http://www.naturalchild.com/elliott_barker
/prisons.htmlHey, stooopid troll Eleaticus - Do you want
EVIDENCE? Each of the 24GPS satellites carries either four
cesium atomic clocks or threerubidum atomic clocks in orbit,
with full relativistic correctionsbeing
applied.http://arXiv.org/abs/hep-th/0307140 GR structure,
especially Part 4/p.
7<http://rattler.cameron.edu/EMIS/journals/LRG/Articles/
Volume4/2001-4will/index.html Experimental constraints on
General
Relativity.<http://tycho.usno.navy.mil/ptti/ptti2002/paper20.
pdfhttp://www.eftaylor.com/pub/projecta.pdf Relativity in the
GPS systemhttp://arXiv.org/abs/gr-qc/9909014 falling
lighthttp://www.hawaii.edu/suremath/SRtwinParadox.html<http://
physics.syr.edu/courses/modules/LIGHTCONE/twins.html Twin
Paradoxhttp://arXiv.org/abs/astro-ph/0401086http://arxiv.org/
abs/astro-ph/0312071 Deeply relativistic neutron star
binarieshttp://arXiv.org/abs/gr-qc/0301024 Nordtvedt EffectNIM
A 355 537 (1995)Physics Letters B 328 103 (1994)Physical Review
Letters 64 1697 (1990)Physical Review Letters 39 1051
(1977)Physical Review 135 B1071 (1964)Physics Letters 12 260
(1964)Europhysics Letters 56(2) 170-174 (2001)General
Relativity and Gravitation 34(9) 1371
(2002)http://fourmilab.to/etexts/einstein/specrel/specrel.pdf<
http://www.geocities.com/physics_world/sr/ae_1905_error.htm<
http://www.physics.gatech.edu/people/faculty/finkelstein/
relativity.pdfhttp://users.powernet.co.uk/bearsoft/Paper6.
pdfhttp://users.powernet.co.uk/bearsoft/LPHrel.html
Longitudinal and transverse
masshttp://www.navcen.uscg.gov/pubs/gps/gpsuser/
gpsuser.pdfhttp://www.navcen.uscg.gov/pubs/gps/sigspec/
default.htmhttp://www.navcen.uscg.gov/pubs/gps/icd200/
default.htmhttp://www.trimble.com/gps/index.htmlhttp://
sirius.chinalake.navy.mil/satpred/http://www.phys.lsu.edu/mog/
mog9/node9.htmlhttp://egtphysics.net/GPS/RelGPS.htmhttp://
www.schriever.af.mil/gps/Current/current.oa1http://
edu-observatory.org/gps/gps_books.html<http://
www-astronomy.mps.ohio-state.edu/~pogge/Ast162/Unit5/gps.html-
-Uncle
Alhttp://www.mazepath.com/uncleal/qz.pdfhttp://
www.mazepath.com/uncleal/eotvos.htm (Do something naughty to
physics)Supersedes:
<physics-faq/criticism/lorentz-intervals_1078308364@
rtfm.mit.eduX-Last-Updated: 1999/10/17===
===
Subject:
: (SR) Lorentz
t', x' = IntervalsSummary: The Lorentz transforms themselves
are proof t' and x' cannot possibly be just coordinates.
Examination of their derivation verifies their identity as
intervals.Originator: faqserv@penguin-lust.MIT.EDUDisclaimer:
approval for *.answers is based on form, not content.
Opponents should first actually find out what the content is,
then think, then request/submit-to arbitration by the
appropriate neutral mathematics authorities. Flaming the hard-
working, selfless, *.answers moderators evidences ignorance and
atrocious netiquette.Version: 0.02.1Archive-name:
physics-faq/criticism/lorentz-intervalsPosting-frequency: 15
days (SR) Lorentz t', x' = Intervals (c) Eleaticus/Oren C.
Webster
Thnktank@concentric.net------------------------------===

===
Subject:
: 1. Introduction with the obvious debunking of the use
of 'just coordinates' in any scientific formula.Defenders of
the Special Relativity faith are especiallyfond of telling
opponents of their space-time fairy talesthat they do not know
the difference between coordinatesand magnitudes. That may
often be so, but the fault lies ultimately with SR dogma. The
Lorentz-Einstein transformations cannot possibly be 'just
coordinates', which is the interpre-tation required to support
the many sideshow carnival actswith which the SR faithful
bedazzle the public, and establishtheir moral and intellectual
superiority.If I get in my car and drive steadily for a few
hours at 50 kilometers per hour, is 50t the distance I travel?
Of course not. The last time my hours-counting 'just
coord-inates' clock was set to zero was when Zeno first
reported one of his paradoxes to Parmenides. That was a long
time ago, so my t is not useful for such purposes unless you
also use my clock to established the starting time, perhaps
t0, and use the formula 50(t-t0) to calculate the distance. In
any case, my t is even then not 'just a coordinate' becauseit
always represents particular elapsed times that can beused in
the (t-t0) form to calculate perfectly good timeintervals
(elapsed times).Alternatively, I could (re)set my clock to
zero at the startof some meaningful time interval, in which
case my t shows a scientifically perfect current and/or end
time. In which case, the Lorentz-Einstein t'=(t-vx/cc)/g is a
function of an elapsed time interval (not 'just a coordinate')
and a time interval (-vx/cc; the interval amount the t' clock
is being screwed up at time t) and thus cannot be 'just a
coordinate' since neither of the independent variables is such
a 'just' thing. {Their meaning is shown below, step-by-step.]If
it takes me 50 minutes to cross the Interstate highway,was x/50
my velocity crossing it?Of course not. The origin of all my
axes is at the veryspot where Zeno first presented his first
paradox to Parmenides. That makes my x equal a couple of
thousands ofmiles, plus, and is not useful for such purposes
unless you establish the starting x value, perhaps x0, and use
theformula (x-x0)/50 to calculate my velocity. In any case,
even then my x is not 'just a coordinate' because it always
repesents particular distance intervalsthat can always be used
in the (x-x0) form for any and everyscientific
purose.Alternatively, I could move my x-axis origin to the
starting(zero) point of some meaningful distance, in which
case my x shows a scientifically perfect current and/or end
distance.In which case, the Lorentz-Einstein x'=(x-vt)/g is a
function of a current/ending distance interval (not 'just a
coordinate') and a distance interval (-vt; the interval amount
the x' axisis being screwed up at time t) and thus cannot be
'just a coordinate' since neither of the independent variables
is such a 'just' thing. {Their meaning is shown below,
step-by-step.]------------------------------===
===
Subject:
: 2.
Table of Contents 1. Introduction with the obvious debunking
of the use of 'just coordinates' in any scientific formula. 2.
Table of Contents. 3. The Lorentz-Einstein transforms. 4. The
'just coordinates' argument. 5. Single-system, little-purpose
ambiguity. 6. Relating two coordinate measures/systems. 7.
Distances and moving coordinate axes. 8. Time intervals. 9.
Einstein's (1905) derivations. 10. A word about intervals. 11.
Intervals versus the Twins Paradox. 12.
Summary------------------------------===
===
Subject:
: 3. The
Lorentz-Einstein transformsSpecial Relativity's space-time
circus is based onthe 'transformation' equations by which it
is believedone can relate a nominally 'stationary' system's
spaceand time coordinates to those of an inertially
(notaccelerating) moving other observer. That moving
observer's own physical body and coordinate system might have
been identical in size to those of the stationary observer
before the traveller began moving, but are 'seen' as very
different by the stationary observer when the relative
velocity of the two is great enough, a high percentage of the
velocity of light.Concerning ourselves - as is customary -
with justthe spatial coordinate axis that lies parallel tothe
direction of motion, and with time, Einsteinarrived at these
formulas that relate the movingsystem measures or coordinates
(x' and t') to thestationary system coordinates (x and t): x'
= (x - vt)/sqrt(1-vv/cc) (Eq 1x) t' = (t -
vx/cc)/sqrt(1-vv/cc) (Eq 1t)The v is for the two systems'
relative velocity as seen by the stationary observer, and is
positive if the dir-ection is toward higher values of x. By
concensus,the moving system x'-axis higher values also lie
inthat direction, and all axes parallel the other
system'scorresponding axis.We used vv to mean the square of v
but might use v^2for that purpose below. Similarly for
c.Because it is believed that no physical object canreach or
exceed c, the square-root term in bothdenominators is presumed
always less than one, which means that the formulas say both x'
and t' will tend tobe greater than x and t, respectively.
However,SRians call the x' result 'contraction' - which
meansshortening - and the t' result 'dilation' - whichmeans
increasing. ------------------------------===
===
Subject:
: 4. The
'just coordinates' argumentThe 'just coordinates' argument is
so patently ridiculousthat even opponents have a hard time
accepting just howsimple and obvious its debunking can be, as
shown in thissection. However, further sections take a more
arithmet-ical approach that you'll maybe find more
professorial.The 'just coordinates' argument is that t is mot
aduration, not a time interval; it's just an arbitraryclock
reading. But what if the moving system observercomes speeding
by while you make your annual 'springforward' or 'fall back'
change? The formula says thatthe moving system clock's 'just
coordinate' reading can be calculated from yours: t' = (t -
vx/cc)/sqrt(1-vv/cc) (Eq 1t)Imagine the moving system
oberver's confusion if his clock changes its reading while
he's looking at it! If his clock doesn't change when yours
does, the formula is wrong; if it is truly a 'just
coordinates' formula. And then what happens if you realize you
were a day early and put your clock back to what it had said
previously?And suppose you are in NYC and your twin in LA
andboth are watching the moving observer. You'll both beusing
the same v because you are at rest wrt (withrespect to) each
other. You're on Eastern StandardTime and your twin is on
Pacific Standard Timemaybe. You have three hours more on your
clock than does your twin. On which 'just coordinate' clock
will the Lorentz transforms base the 'just coordinate' time
the moving system clock says? The formula applies to both
ofyour t-times: t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq 1t)Sure,
the idea that you can change someone else'sclock with no
connection of any kind is really ridiculous, but Eqs 1x and 1t
aren't MY equations. Are they yours? And we aren't the ones to
say x, t, x', and t' are just coordinates.If the t' formula is
actually either an elapsedtime formula, or the basis of a t'/t
ratio, thenthere is no implication that one clock's readinghas
anything to do with the other's. It can only be rates of clock
ticking, or how one time INTERVAL compares to the other that
the formula is about.------------------------------===
===
Subject:
:
5. Single-system, little-purpose ambiguity.Since we're going to
be comparing measurements on twocoordinate systems in the next
section, let's go toour supply cabinet and get our yard-stick
(which weuse to measure things in inches) and our
meter-stick(which we use to measure things in
centimeters).Here, I'm getting mine. Oh! Oh!There's an ant on
mine, and he ... she ... sure ishanging on, right at the 3.5
inch mark of the yard-stick.Let's see if I can wave the stick
around enough thatshe'll let go. Nope.However, before I gave
up I waved the stick and theant 'all over the place.Always,
however, the ant was at the 3.5 mark on the yard-stick, and
always 3.5 away from the end of thestick, however far and wide
I have transported her.Neither of those 3.5 facts means very
much. Of thetwo, the distance aspect meant almost nothing.
Sothe distance was 3.5 from the end. So what? Thatlength,
distance, was not in use. And only maybethe ant might have
been concerned with just whatlocation, 'just coordinate', on
the stick she wasat.Just so with x and t.So, is the 3.5
reading just a coordinate? Or adistance/length? It's ambiguous
in and of itself,and really makes no difference what you say
untilyou try to make use of the number. Hey, my address is
5047 Newton Street. If youare looking for me and you're at
4120 Newton, itis helpful information, because it tells you
whichdirection to go. Is that 'just coordinate'? Where it
really becomes useful, perhaps, is intelling you how far away
I am. That's not justa coordinate value, that's a distance,
length,interval.However, it is subtracting 4120 from 5047 that
tells you which direction and how far. It is only because both
5047 and 4120 are distances from the same point - ANY same
point - that the result means anything.My x - my yardstick
reading - is always a distanceor length; it is impossible to
be otherwise withan honest, competently designed
yardstick.Whether or not its reading is of good use in some
particular scientific formula depends on whether I put the
zero end of the yardstick at some usefulplace. As in the
introduction, we should eitherput it at the starting
location/end, or use tworeadings from it:
(x-x0).------------------------------===
===
Subject:
: 6. Relating
two coordinate measures/systems.Taking care to not damage our
brave little ant, I placemy yard-stick onto the table, zero
end to the left, 36end to the right.Now I place the 'just
coordinate' meter-stick on the tablein the same orientation,
in a random location, and findthat the ant's coordinate on the
meter-stick is 51.The formula relating centimeters to inches is
cm=i*2.54but we want a formula similar to
x'=(x-vt)/sqrt(1-vv/cc).That would be c=i/.03937
approximately, but let's use x'for the meter-stick reading,
and x for the inch reading: x'=x/.3937. 3.5/.3937 = 8.89 Wait
a minute. It's not just science but definition that says
c=i/.3937=8.89, so something is wrong. 8.89is not 51.We
already knew that 51 cm was just an arbitrary coordinate.
Arbitrary not because that point isn't 51 cm from the zero end
of the meter-stick, but because the zero point was in an
arbitrary position.Let's put the meter-stick in a position
where it's zero point is at the yard-stick zero point.What is
the centimeter coordinate now? Hey. 8.89,just like the formula
says.The only way for a 'transform' like x'=x/g to work,
whatever g might be, is for both coordinate systemsto have
their zero points aligned, in which casesaying the two
measures are not intervals is pureidiocy.Noe that with both
zero points at the same positionboth x' and x are great
measures for scientificpurposes, in any and every case where
we were smartenough to put those zero points at a useful
location.There is one extension of x'=x/g that will let ususe
the meter-stick in arbitrary position. When the cm reading was
51, the zero point of theyard-stick read (51-8.89=) 42.11 cm.
If we call thatpoint x.z' we get x' = x.z' + x/.3937. = 42.11
+ 3.5/.3937 = 42.11 + 8.89 = 51.Obviously, in this formula
x/.3937 is the distancefrom the x' coordinate of the location
where x=0. An interval.Just as obviously, the fact that we now
have thecorrect formula for relating an x interval to
anarbitrary x' coordinate, does not mean that x'is anything
more than nonsense for use in anyscientific formula.Unless we
were smart enough to put the x zeropoint in a useful location,
and use (x'-x.z') inthe scientific formula. (x'-x.z') equals
the useful,Ratio Scale value x/.3937.So, we have discovered a
basic fact: a transformationformula like x'=x/g works only if
the two zero pointsof the coordinate systems coincide. That
makes it non-sense to say the two coodinates are only
coordinatesand not intervals. Both must be values that
representdistances from their respective zero points unless
youtake the proper steps to adjust for the discrepancy.Make
sure you understand that although the inclusionof x.z' made it
possible to correctly calculate x',the result is nonsense when
it comes to use of x'for general length/distance purposes; it
is x'-x.z' that is a useful number in such cases. It could
bethat we're measuring a sheet of paper with one endat x=0 and
the other at x=3.5; x'=51 is nonsense asa centimeter measure of
the paper.But, you say, the Lorentz transform contain a -vt
term.------------------------------===
===
Subject:
: 7. Distances
and moving coordinate axes.We discovered x'=x.z' + x/g as the
correct formulafor relating one coordinate to another
system's.But the Lorentz transform contains another term,
-vt/sqrt(1-vv/cc). What is it?Let's start with our x'=51 cm,
x=3.5, x.z'=42.11 example.Every minute, let's move the
meter-stick one inch to ourright.At minute 0, the cm reading
was 51 cm.At minute 1, the cm reading is now 50 cm.At minute
2, the cm reading is now 49 cm.In this instance, v=1
inch/minute. And t was 0, 1, 2.What has happened is that we
have made our x.z' a lie,and increasingly so. -vt/.3937 is the
change in x.z'. x' = (x.z - vt/.3937) + x/.3937.Obviously,
vt/.3937 is not a coordinate; even most SRianswouldn't imagine
it was. It is an interval, the distanceover which the moving
system has moved since t=0.And, of course, x/.3937 is the
distance of our bravelittle ant from the point where x=0 and
the centimeterreading is x.z'-vt/.3937. Yes, every minute the
meter-stick moves to the right and the meter-stick
coordinateof the spot where x=0 gets less and less - and
eventuallynegative.Make sure you understand that every minute
the x' coordinate, because of -vt/g, becomes a better measure
of, say, the 3.5 paper we might be measuring with the
yard-stick, given that 51 was too big a number and-vt is
negative. That is, until the two origins coincide at x'=x=0,
and then it gets worse and worse.With -vt positive (because
v<0) the situation is different.With 51 and -vt positive, x'
just gets worse and worseover time.Quite obviously, the fact
that we now have thecorrect formula for relating an x interval
to anarbitrary x' coordinate even when the x' axis ismoving,
does not mean that x'is anything more than nonsense for use in
any scientific formula.Unless we were smart enough to put the x
zero point in a useful location, and use (x'-x.z'+vt/.3937)
inthe scientific formula. (x'-x.z'+vt/.3937) equals the
useful, Ratio Scale value
x/.3937.------------------------------===
===
Subject:
: 8. Time
intervals.Instead of using our sticks, let's get out two
clocks.Mind you, we're not going to deal with different
clockrates here, just establish the same basics as for
distance.Your clock says 9:00 Eastern Standard Time (EST) and
we note that t=540 minutes when we put down the clock.Blindly,
let's turn the setting knob of your twin's Pacific Standard
Time clock and put it down before us.According to what we see,
EST's 540 minutes (9:00) corre-sponds to PST's 14:30; t'=870.We
know the formula relating PST to EST is t' (pacific)= t
(eastern) - 180 (minutes). Thus, it is not correct that the
second clock can have an arbitrary setting, because 870 <>
540-180. We know that the two clocks are related by t' = t/1
since both are using the same second, hour, etc units. But 870
(14:30 in minutes) is not 540/1-180, so once again we know
something is wrong. However, t'=t.z' + t/1 works. EST midnight
equals PST 0.0 (midnite) - 180, so t.z' = -180, and t' = -180 +
540/1 = 360.Since EST-180=PST, 9:00 EST is 6:00 PST = 360
minutes.We see thus that like distance measures/coordinates,
timeaxis origins (zero points) must either be 'lined up' or
adjusted for. So, the Lorentz/Einstein t'=t/sqrt(1-vv/cc) must
be the moving system elapsed time interval since the time axes
were both at a common zero. There is no t.z' adjustment: t' =
(t - vx/cc)/sqrt(1-vv/cc) (Eq 1t)Make sure you understand that
in the clock case, if theEST is showing a good number for
elapsed time since thetravelling observer passed NYC, then the
PST clock issilliness. t.z' must be zero or must be taken out
oftime lapse calculations for the PST clock to be
usedintelligently, just as was true for x.z'.What is lacking
as yet for Lorentz t' is the -vx/cc term thatcorresponds to
the x' formula -vt term.Break it up into two parts: v/c and
x/c. v/c is a scaling factor that changes velocity from
whatever kind of unit you are using over to fractions of c.x/c
is distance divided by velocity, which is time. x/cis thus the
time interval since the two time axeshad a common zero point -
which they have to have in theLorentz transforms which do not
have the t.z' term welearned to use above.Thus,
(-vx/cc)/sqrt(1-vv/cc) is the interval amount the moving
system clock has been changed - since the common/adjusted time
- over and beyond the elapsed time intervalrepresented by
x/sqrt(1-vv/cc).We have discovered that the only way for t' to
be t/gis for t' and t to have a common zero point, just asfor
x' and x. It would be otherwise if the t' formulacontained an
adjustment t.z' under some name or other,but the necessity to
include such a term correlates100% with t' numbers that aren't
directly usable.As for x and x', our knowledge of how to setup
a properformula relating t and t' is of no use unless we
usethe knowledge in scientific formulas; (t'-t.z'+xv/gcc)gives
us the only directly useful value:
t/g.------------------------------===
===
Subject:
: 9. Einstein's
(1905) derivations.When we return to Einstein's derivations of
the transformformulas with a well-focused eye, we find he was a
wee bitconfused - or at least self-contradictory.When he set up
his (at first unknown) tau=moving systemtime formulas, he
created three particular instances of tau.Tau.0 is the time at
which light is emitted at the movingorigin toward a mirror to
the right that is moving at rest wrt that moving origin and at
a constant distance from that origin. He lets the stationary
time slot have the value t,a constant, the stationary system
starting time.Tau.1 is the time at which the light is
reflected. Helets the stationary time be t+x'/(c-v); t is
still aconstant and x'/(c-v) is the time interval since
t.Tau.2 is the time at which the light gets (back) to
themoving origin. The stationary time value is put as t
+x'/(c-v) + x'/(c+v); t is still a constant and x'/(c-v)+
x'/(c+v) is the time interval since t.On the thesis that the
moving observer sees the time tothe mirror as the same as the
time back to the origin,he sets .5[ tau.0 + tau.2 ] =
tau.1.Tau.0 completely drops out of the analysis and leavesno
trace, and has no effect.Further, the t you see in tau.0,
tau.1, and tau.2 also completely drops out with no trace and
no effect, leaving us with exactly what you'd get if you had
explicilty said t' is an interval and so is t.What doesn't
drop out in the stationary time values isx'/(c-v) and
x'/(c+v), the time interval it takes forlight to get to the
fleeing mirror, and the time intervalit takes for light to get
back to the approaching origin.Thus, his resultant t' formula
is strictly based on time intervals in the stationary system.
Time intervals since some starting time, yes, but time
intervals.There is absolutely nothing in the derived formulas
that depends on arbitrary coordinates like the constant t in
the stationary time arguments.Let's look at the x dimension;
it is x'=x-vt [as x increasesby vt, the effect over time is
x'=(x+vt)-vt)], which Einstein explicitly sets up as a
constant stationary distance.He uses that x' not just in the
time interval parts of the stationary time arguments, but also
in the x (distance) stationary system argument for the tau at
the time light is reflected. x' can't be the stationary system
coordinate of the mirror at that time. That value is x'+vt.x'
is explicitly an interval, distance. Thus, the whole tau
derivation of the t' formula is fully andexplicitly based on
x' - a spatial length/distance/interval -and the two time
interals x'/(c-v) and x'/(c+v).While we're at it, if the
starting t is not zero, his x'=x-vt formula is complete
nonsense also. Given thatthere was some L that was the mirror
x-location and lengthwhen the light is emitted, if t was
already, say, 500, thenx'=L-vt could have been a very negative
length.------------------------------===
===
Subject:
: 10. A word
about intervals.There are intervals, and there are
intervals.If we put our yard stick zero point at one endof a
piece of paper and read off the coordinateat the other end of
the paper, we have a goodmeasure of the paper's length, a
Ratio Scalemeasure. [Absolute temperature scales are
ratioscale.]If instead we put the one end of the paper at
theone inch mark (or the zero end of the stick oneinch 'into'
the length of the paper) we get measuresthat are one inch off
the true, ratio scale length.The two messed up measures are
still intervals,but they are Interval Scale measures.
[Householdtemperature scales are interval scale, which iswhy
your physics and chemistry professors won'tlet you use them
without first converting to theratio scale absolute
temperatures.)t'=t/g and x'=x/g represent ratio scale
measures,given that t and x were ratio scalae to start
with.t'=t.z'+t/g and t'=t/g-vx/gcc are both interval scale
measures, even given a good ratio scale tand a good ratio
scale x.x'=x.z'+x/g and x'=x/g-vt/g are both interval scale
measures, even given a good ratio scale xand a good ratio
scale t.Look for the (SR) Lorentz t', x' = degraded
measuresdocument soon at a newsgroup near
you.------------------------------===
===
Subject:
: 11. Intervals
versus the Twins Paradox.t'=(t-vx/cc)/g shows t' being greater
than t.The reason Special Relativity will not allow theuse of
its basic time equation in determining whatSR has to say about
the twins' ages, is that t' andx' are supposedly just
coordinates, and they say you have to take the coordinate
pairs (t',x') and (x,t)into consideration in both the time and
place the twins' separation started and the time and place the
twins reunited.Since t' and x' are actually both intervals,
notjust coordinates, the 'excuse' is spurious, and is so even
without use of the obvious (x_b-x_a) and(t_b-t_a)
usages.However, SR is right to be embarrassed by
theirtransformation formulas.Look for the (SR) Lorentz t', x'
= degraded measuresdocument at a newsgroup near
you.------------------------------===
===
Subject:
: 12. SummaryA.
t'=t/g and x'=x/g can be almost 'just coordinates' in the
sense that the values obtained may not be of much use except
in the most primal and useless way: how long and how far
since/from the time/ place they were zero. Even here, however,
the zero points within each of the two scale pairs (t',t) and
(x'.x) must have been lined up. If the zero points have been
intelligently selected (such as at the starting point and time
of a trip) they can be rationally used 'as is' in any valid
sci- entific equation.B. Even the interval scale t'=t.z' -
xv/gcc + t/g and x'=x.z' - vt/g + x/g are not 'just
coordinates'. They can be used to good effect by establishing
the relevant starting times/points and using (t'-t.z'+xv/gcc)
and (x'-x.z'+vt/g), as the situation may require.C. When you
see vx/gcc or vt/g in use in any guise with non-zero values,
you know the resultant t' or x' is a degraded, interval scale
value.E-X: Anytime you do not see what amounts to t.z' and
xv/gcc in the time case, or x.z' and vt/g in the distance
case, you know that the t' and/or x' in use are intervals.
Period.Y: Either set your clock to zero at the start of the
relevant time interval, or use (t-t0), with both being
readings on the same clock. Either move your x-axis origin to
the starting end or point, or use (x-x0), with both being
readings on the same axis.Z: In _(SR) Lorentz t', x' =
Degraded (Interval) Scales_ we see that t' and x' satisfy the
mathematical tests for/of interval scales when -vt and -vx/cc
are not zero; thus, they must be intervals. When -vt and
-vx/cc are zero, t' and x' satisfy the much better
mathematical definition of ratio scales, and are thus not just
mere intervals, but (rescaled) good
ones.Eleaticus!---?---!---?---!---?---!---?---!---?---!---?---
!---?---!---?---!---?! Eleaticus Oren C. Webster
ThnkTank@concentric.net ?! Anything and everything that
requires or encourages systematic ?! examination of premises,
logic, and conclusions
?!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
---!---?===
===
Subject:
: Re: (SR) Lorentz t', x' = Intervals> (SR)
Lorentz t', x' = Intervals> (c) Eleaticus/Oren C. Webster>
Thnktank@concentric.net[snip 1300 lines of trolled
garbage]Originally trolled across sci.physics
sci.physics.relativityalt.physics sci.math sci.answers
alt.answers
news.answershttp://b5.sdvc.uwyo.edu/bab5/snds/
argcstpd.wavPsychotic ineducable boring troll Eleaticus, You
see yourself this
way,http://www.mazepath.com/uncleal/effete6.jpg The entire
remainder of the planet sees you this
way,http://www.mazepath.com/uncleal/effete3.pnghttp://
w0rli.home.att.net/youare.swfhttp://www.mazepath.com/uncleal/
sunshine.jpghttp://www.you-moron.com/http://www.apa.org/
journals/psp/psp7761121.htmlhttp://insti.physics.sunysb.edu/~
siegel/quack.html<http://www.firehead.org/~jessh/film/kubrick/
Kubrick-Psycho.html<http://www.naturalchild.com/elliott_barker
/prisons.htmlHey, stooopid troll Eleaticus - Do you want
EVIDENCE? Each of the 24GPS satellites carries either four
cesium atomic clocks or threerubidum atomic clocks in orbit,
with full relativistic correctionsbeing
applied.http://arXiv.org/abs/hep-th/0307140 GR structure,
especially Part 4/p.
7<http://rattler.cameron.edu/EMIS/journals/LRG/Articles/
Volume4/2001-4will/index.html Experimental constraints on
General
Relativity.<http://tycho.usno.navy.mil/ptti/ptti2002/paper20.
pdfhttp://www.eftaylor.com/pub/projecta.pdf Relativity in the
GPS systemhttp://arXiv.org/abs/gr-qc/9909014 falling
lighthttp://www.hawaii.edu/suremath/SRtwinParadox.html<http://
physics.syr.edu/courses/modules/LIGHTCONE/twins.html Twin
Paradoxhttp://arXiv.org/abs/astro-ph/0401086http://arxiv.org/
abs/astro-ph/0312071 Deeply relativistic neutron star
binarieshttp://arXiv.org/abs/gr-qc/0301024 Nordtvedt EffectNIM
A 355 537 (1995)Physics Letters B 328 103 (1994)Physical Review
Letters 64 1697 (1990)Physical Review Letters 39 1051
(1977)Physical Review 135 B1071 (1964)Physics Letters 12 260
(1964)Europhysics Letters 56(2) 170-174 (2001)General
Relativity and Gravitation 34(9) 1371
(2002)http://fourmilab.to/etexts/einstein/specrel/specrel.pdf<
http://www.geocities.com/physics_world/sr/ae_1905_error.htm<
http://www.physics.gatech.edu/people/faculty/finkelstein/
relativity.pdfhttp://users.powernet.co.uk/bearsoft/Paper6.
pdfhttp://users.powernet.co.uk/bearsoft/LPHrel.html
Longitudinal and transverse
masshttp://www.navcen.uscg.gov/pubs/gps/gpsuser/
gpsuser.pdfhttp://www.navcen.uscg.gov/pubs/gps/sigspec/
default.htmhttp://www.navcen.uscg.gov/pubs/gps/icd200/
default.htmhttp://www.trimble.com/gps/index.htmlhttp://
sirius.chinalake.navy.mil/satpred/http://www.phys.lsu.edu/mog/
mog9/node9.htmlhttp://egtphysics.net/GPS/RelGPS.htmhttp://
www.schriever.af.mil/gps/Current/current.oa1http://
edu-observatory.org/gps/gps_books.html<http://
www-astronomy.mps.ohio-state.edu/~pogge/Ast162/Unit5/gps.html-
-Uncle
Alhttp://www.mazepath.com/uncleal/qz.pdfhttp://
www.mazepath.com/uncleal/eotvos.htm (Do something naughty to
physics)===
===
Subject:
: Re: anthyphairetic ratio> Mathworld
says:An archaic word for a continued fraction is
anthyphairetic ratio.There's a similar word Prosthaphairesis
[sp?], which means the useof half-angle trig identities to
perform multiplication. This was astandard technique in the
middle ages, before the discovery of logs.CheersJohn Ramsden
(`rm -rf /`@.sod_off_spammers.com)===
===
Subject:
: Re:
anthyphairetic ratio charset=utf-8 John Ramsden
<john_ramsden@sagitta-ps.com> [CapitalEth][EDoubleDot][Micro]
.b3 .b9[EDoubleDot].b9> There's a similar word
Prosthaphairesis [sp?], which means the use> of half-angle
trig identities to perform multiplication. This was a>
standard technique in the middle ages, before the discovery of
logs.The difference being that prosthaphairesis has survived
into modern Greekand means any sequence (in any order) of
successive additions/subtractions,while anthyphairesis hasn't
survived. At least as far as I know.> Cheers> John Ramsden
(`rm -rf /`@.sod_off_spammers.com)--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
understandable===
===
Subject:
: Re: Hermite functions: Why discrete
eigenvalues?[...]> Well, for problem (1) the explanation is
similar. When |x| is large we maydisregard lambda, and hence
we get the simpler expression: D^2 u = x^2 u.> Since the
coefficient to u is positive here, we have exponential-like>
behaviour (as opposed to oscillatory behaviour). So in general
the solutions> will tend to grow unboundedly, unless the
conditions are exceptional. These> correspond exactly to the
eigenvalues.Yeah, I have the same feeling, namely, that the
solution would convergeonly in exceptional cases. But, my
problem is, what are those exceptionalcases? why and how are
they exceptional?[...]> Try finding a solution with the
additional conditions u(0) = 1 and u'(0) => 0. Then see what
happens as x approaches +infinity, for various values of>
lambda. It will oscillate when x^2 < lambda, and do the
opposite (I can't> find the appropriate word to use here) when
x^2 > lambda.That's a good idea! If we regard x as time and u
as position, then we areu(0) = 1 and u'(0) = 0. . . . Now I
begin to see. . . Starting frombecause there is a restoring
force -lambda u. But, as time goes on, thebecomes larger and
larger. When t = sqrt(lambda), those two forces finallymatch,
and after that the acceleration t^2 u is stronger. Then,
theits velocity and position at t = sqrt(lambda) are right.
First, the velocitymust be in the direction of the origin.
Then there's a hope: even though theacceleration t^2 u is
proportional to t^2, it can remain weaker than therestoring
force if u approches to the origin fast enough. But, the
velocityovershoot the origin and then it can never come back
(because it can'toscillate any longer). So, the value of
lambda must be such as gives thecorrect velocity at t =
sqrt(lambda).This is an appealing explanation. I'm not still
sure why lambda must bediscrete. It could be that the velocity
is right for a continuous rangeof lambda. . . .Anyway, your
explanation is illuminating. Thank you, Michael.Ryo>
-Michael.===
===
Subject:
: very easy topology problem...empty set is
compact.....----------------------um.....i want to know the
reason===
===
Subject:
: Re: very easy topology problem...> empty set
is compact.....> ----------------------> um.....i want to know
the reasonTake an open covering of the empty set. Let A be
anyelement of the covering. Then A contains the empty setand
so {A} is a finite sub-covering.Of course, you can also take
the empty subset of thecovering...Jose Carlos
Santos===
===
Subject:
: Re: very easy topology problem...>> empty
set is compact.....>Take an open covering of the empty set.
Let A be any>element of the covering. Then A contains the
empty set>and so {A} is a finite sub-covering.This works
unless you consider the empty set as an open cover for
theempty set.-- I'm not interested in mathematics that might
have anythingto do with reality. -- Easterly, in
sci.math===
===
Subject:
: Re: very easy topology problem...> empty
set is compact.....>>Take an open covering of the empty set.
Let A be any>>element of the covering. Then A contains the
empty set>>and so {A} is a finite sub-covering.>This works
unless you consider the empty set as an open cover for
the>empty set.But the empty set is an open cover for the empty
set. So it doesn't work.Derek Holt.===
===
Subject:
: easy....topology
problemclosure of compact set is compactfind counter
example.--------------------------um....help me,
please.....===
===
Subject:
: Re: easy....topology problem> closure
of compact set is compact> find counter example.The include
point topology.Topology = { U | p in U }Thus { p } is compact
and cl { p } = whole space which is compact iff it's
finite.===
===
Subject:
: easy...easy...topology problem...let T and
T' be two topologies on the set X.suppose that T' < T1. if
(X,T) is compact , then (X,T') is compact2. if (X,T') is
compact , then (X,T) is compacttrue or
false??----------------------------------------------i
think.......1. true.2. falsei am right??===
===
Subject:
: Re:
easy...easy...topology problem...Try this :Post hoc, ergo
propter hoc.> let T and T' be two topologies on the set X.>
suppose that T' < T> 1. if (X,T) is compact , then (X,T') is
compact> 2. if (X,T') is compact , then (X,T) is compact> true
or false??> ----------------------------------------------> i
think.......> 1. true.> 2. false> i am right??===
===
Subject:
: Re:
easy...easy...topology problem...> let T and T' be two
topologies on the set X.> suppose that T' < T> 1. if (X,T) is
compact , then (X,T') is compact> 2. if (X,T') is compact ,
then (X,T) is compact> true or false??>
----------------------------------------------> i
think.......> 1. true.> 2. false> i am right??Yes.Jose Carlos
Santos===
===
Subject:
: Re: Beall Conjecture> On June 11, 2000, I
posted a brief solution to the Beall conjecture. > There is a
good bit of money riding on the proof. Perhaps my post> merits
a bit more attentionDid you submit your claimed solution to the
American MathematicalSociety for vetting. If your proof holds
up you should collect $100,000.===
===
Subject:
: Re: Moduleseach of
these cases? What ring is the set of vector bundle sections
amodule over? Or a representation?from Jonny!> |Hi!> |> |I've
been studying some module theory and I particularly like how
its> |framework is sufficiently general to cover both the
classification of> |abelian groups and the normal forms of
matrices. Aside from the> |classic examples (abelian groups as
Z-modules, vector spaces as F[x]> |modules with mutliplication
defined by f(x).v = f(a)(v) for some a in> |End(V)) are there
any more interesting examples of modules?> there are lots of
them. for example, representations of groups or of> lie
algebras, and the collection of sections of a vector bundle.
(if> sections of a vector bundle isn't familiar to you yet,
then don't> worry; it's not that hard to learn about. same for
group> representations and lie algebras.)> |Maybe ones> |which
provide a new way of looking at a familiar problem, making it>
|easier to understand?> yes, this happens a lot, though
sometimes blended together withproviding a way of becoming
familiar with a previusly not-so-familiar> problem.===
===
Subject:
:
Re: Modules|each of these cases? What ring is the set of vector
bundle sections a|module over?over a ring of functions on the
base space over which the vectorbundle lies. there's actually
a number of variations on exactly howthis works, depending on
whether the vector bundle and the base spacehave topological
structure, or smooth structure, or analyticstructure, or
whatever.for example, if the base space is a topological space
and the vectorbundle over it is a topological bundle of real
vector spaces, then themodule is best taken to be the set of
_continuous_ sections of thevector bundle, and the ring to be
the set of _continuous_ functionsfrom the base space to the
real numbers.the module structure is assigned like this: given
a section s of thevector bundle and a function f on the base
space, the new section f*sis assigned to have the same value
as s does at any point x of thebase space, except
scalar-multiplied by the number f(x).thinking about this
example and about the original two examples thatyou mentioned
leads to the pretty important idea that modules over
acommutative ring are sort of like families of vector
spacesparameterized by the spectrum of the commutative
ring.|Or a representation?a representation is a module of the
group ring of the group that'sbeing represented. again there
are minor variations depending onstuff like what field the
representation is a vector space over; sofor example if v is a
complex vector space on which the group g isrepresented, then v
is also a module of the group ring of g withcomplex
coefficients.the module structure is assigned in a pretty
straightforward way,extending by linearity. that is, an
element f of the group ring ofg with coefficients in the field
k is a linear combination k1*g1+...+kn*gnof elements of g with
coefficients in k, and given an element s of ak-vector space
on which g is represented, the new element f*s is takento be
the corresponding linear combination
k1*[g1*s]+...+kn*[gn*s].|from Jonny!||||> |Hi!|> ||> |I've
been studying some module theory and I particularly like how
its|> |framework is sufficiently general to cover both the
classification of|> |abelian groups and the normal forms of
matrices. Aside from the|> |classic examples (abelian groups
as Z-modules, vector spaces as F[x]|> |modules with
mutliplication defined by f(x).v = f(a)(v) for some a in|>
|End(V)) are there any more interesting examples of
modules?||> there are lots of them. for example,
representations of groups or of|> lie algebras, and the
collection of sections of a vector bundle. (if|> sections of a
vector bundle isn't familiar to you yet, then don't|> worry;
it's not that hard to learn about. same for group|>
representations and lie algebras.)||> |Maybe ones|> |which
provide a new way of looking at a familiar problem, making
it|> |easier to understand?||> yes, this happens a lot, though
sometimes blended together with|providing a way of becoming
familiar with a previusly not-so-familiar|> problem.-- [e-mail
address jdolan@math.ucr.edu]===
===
Subject:
: The miller and the
donkey problem - a reason of a
discrepancyhttp://santinho-de-pau-a-fala.blogspot.com/Manuel
C. Sousa===
===
Subject:
: Re: FIFO problem - yet another .sig rot
script...>Oh, heck, and here I was waiting for someone to tell
you to>pre- and postmultiply your .sig by a unit quaternion.a
unit complex number would have sufficed... never
mind!!;-)Michele-- you'll see that it shouldn't be so. AND,
the writting as usuall isfantastic incompetent. To illustrate,
i quote:- Xah Lee trolling on clpmisc, perl bug File::Basename
and Perl's nature===
===
Subject:
: Re: terminology questions:
partition vs. quotient set, and graph vs. extension> If a
quotient set of S is S/R, where R is an equivalence relation>
on S, then a quotient set of S is exactly the same thing as a>
partition of S - both are a pairwise-disjoint collection of
subsets> of S with union S.Is the redunt terminology then just
due to multiple historicalpaths to the same concept?>2. What's
the difference between the graph of a relation and
the>extension of a relation?> What is the extension of a
relation?>The set of objects which satisfy the relation.>
Presumably meaning the set of ordered pairs (x,y) which>
satisfy the relation? If so, that's exactly what the relation>
is (and also exactly what the graph of the relation is).Redunt
terminology for the same reason here also?===
===
Subject:
: Re:
terminology questions: partition vs. quotient set, and graph
vs. extension>> If a quotient set of S is S/R, where R is an
equivalence relation>> on S, then a quotient set of S is
exactly the same thing as a>> partition of S - both are a
pairwise-disjoint collection of subsets>> of S with union
S.>Is the redunt terminology then just due to multiple
historical>paths to the same concept?>>2. What's the
difference between the graph of a relation and the>>extension
of a relation?> What is the extension of a relation?>>The
set of objects which satisfy the relation.> Presumably meaning
the set of ordered pairs (x,y) which>> satisfy the relation? If
so, that's exactly what the relation>> is (and also exactly
what the graph of the relation is).>Redunt terminology for the
same reason here also?This depends on the context. I was
assuming that we weretalking about the word relation is it's
used in math. Ifyou're talking about relations, as in relation
symbol informal _logic_ then what I said about a relation
beingidentical to its extension is false: in that context
theextension of a relation symbol is the _interpretation_of a
relation in a structure; they're not the same thing at
all.===
===
Subject:
: Re: terminology questions: partition vs.
quotient set, and graph vs. extension> If a quotient set of
S is S/R, where R is an equivalence relation> on S, then a
quotient set of S is exactly the same thing as a> partition
of S - both are a pairwise-disjoint collection of subsets>
of S with union S.> Is the redunt terminology then just due to
multiple historical> paths to the same concept?Or alternate
ways of looking at it. A partition is the set of blocks.A
quotient set is what is obtained when each block is identified
to apoint.>Presumably meaning the set of ordered pairs (x,y)
which>satisfy the relation? If so, that's exactly what the
relation>is (and also exactly what the graph of the relation
is).> Redunt terminology for the same reason here also?Depends
on your foundations. Whether a function is its graph, orperhaps
defined in some other way. (You can set up mathematics,
say,where functions are the fundamental objects, and sets are
specialkinds of functions.) In any case, the graph of a
function is theset of ordered pairs.===
===
Subject:
: Re:
terminology questions: partition vs. quotient set, and graph
vs. extension>1. What's the difference between a partition and
a quotient set?> If a quotient set of S is S/R, where R is an
equivalence relation> on S, then a quotient set of S is
exactly the same thing as a> partition of S - both are a
pairwise-disjoint collection of subsets> of S with union S.>2.
What's the difference between the graph of a relation and
the>extension of a relation?> What is the extension of a
relation?>The set of objects which satisfy the relation.>
Presumably meaning the set of ordered pairs (x,y) which>
satisfy the relation? If so, that's exactly what the relation>
is (and also exactly what the graph of the relation is).Yes,
that's tighter.I was introduced to the concept of extension in
a class on modalsemantics. Kripke's influence was quite strong,
as you can probablysee. :)However, saying that that's exactly
what the relation is doesn'tseem right to me. Presumably,
since we're talking about the extensionof a predicate, we're
doing some model theory. But to say that apredicate *is* the
set of n-tuples which satisfy it seems circular.I'm sure
there's some way out of this trouble, I'm just too hungry
tothink right now.'cid 'ooh===
===
Subject:
: Re: terminology
questions: partition vs. quotient set, and graph vs.
extension>>1. What's the difference between a partition and a
quotient set?> If a quotient set of S is S/R, where R is an
equivalence relation>> on S, then a quotient set of S is
exactly the same thing as a>> partition of S - both are a
pairwise-disjoint collection of subsets>> of S with union
S.>2. What's the difference between the graph of a relation
and the>>extension of a relation?> What is the extension of a
relation?>>The set of objects which satisfy the relation.>
Presumably meaning the set of ordered pairs (x,y) which>>
satisfy the relation? If so, that's exactly what the
relation>> is (and also exactly what the graph of the relation
is).Yes, that's tighter.>I was introduced to the concept of
extension in a class on modal>semantics. Kripke's influence
was quite strong, as you can probably>see. :)>However, saying
that that's exactly what the relation is doesn't>seem right to
me. Presumably, since we're talking about the extension>of a
predicate, we're doing some model theory. But to say that
a>predicate *is* the set of n-tuples which satisfy it seems
circular.I wasn't assuming we were doing model theory. There's
a perfectlystandard notion of relation in mathematics other
than logic; thedefinition of a binary relation is as a certain
set of ordered pairs.Otoh you're probably right about what the
OP had in mind, sincethe notion of extension would not come up
otherwise.>I'm sure there's some way out of this trouble, I'm
just too hungry to>think right now.>'cid 'ooh===
===
Subject:
:
Vector product [Q] (name this matrix)X-No-Confirm: yesWe are
in R^3, with basis e_1, e_2, e_3. Let v be any vector inR^3.
Is there a name for the matrix whose columns are the
vectorse_i x v (i = 1, 2, 3)? (x here refers to the vector
product inR^3, aka cross-product.) I'm looking for theorems
and identitiesfeaturing this matrix, and it would be helpful
to have a name tosearch by.Many thanks in advance,jill-- To
s&e^n]d me m~a}i]l r%e*m?ov[e bit from my
a|d)d:r{e:s]s.===
===
Subject:
: Re: Vector product [Q] (name this
matrix)>We are in R^3, with basis e_1, e_2, e_3. Let v be any
vector in>R^3. Is there a name for the matrix whose columns
are the vectors>e_i x v (i = 1, 2, 3)? (x here refers to the
vector product in>R^3, aka cross-product.) I'm looking for
theorems and identities>featuring this matrix, and it would be
helpful to have a name to>search by.Typically (if not always),
statements about cross product inR^3 are more enlightening
when translated into statements aboutwedge product (or
exterior product). I think your questionis an example. Here is
a rephrasing, with extra padding, inexterior-algebraic terms.We
are in R^3, with dual space (R^3)* and exterior
algebraLambda(R^3), where by construction (or definition)
Lambda(R^3)is a graded algebra whose degree-0 part is
identified with R,whose degree-1 part is (R^3)*, and which is
generated as anexterior algebra by its degree-1 part; and
where, therefore,by definition (or construction) Lambda(R^3)
is the algebraof alternating multilinear forms on R^3. Let
lambda be anycovector in R^3, that is, any element of (R^3)*.
Considerthe transformation mu |--> mu ^ lambda from (R^3)*to
(R^3)* ^ (R^3)* (that is, from the [necessarily
alternating!]1-forms to the alternating 2-forms; that is, from
the degree-1part of the exterior algebra to the degree-2 part).
Thenthis transformation is linear. WHEN ONE USES (for
instance)A METRIC (or a choice of basis, or ...) TO IDENTIFY
R^3 with(R^3)*, and consequently to make many other
identifications ofvarious 3-dimensional vectorspaces, THEN THE
MATRIX OF THISwedge-with-lambda TRANSFORMATION becomes the
matrix you'reinterested in.Lee Rudolph===
===
Subject:
: Re: Vector
product [Q] (name this matrix)X-No-Confirm: yes>>We are in
R^3, with basis e_1, e_2, e_3. Let v be any vector in>>R^3. Is
there a name for the matrix whose columns are the vectors>>e_i
x v (i = 1, 2, 3)? (x here refers to the vector product
in>>R^3, aka cross-product.) I'm looking for theorems and
identities>>featuring this matrix, and it would be helpful to
have a name to>>search by.>Typically (if not always),
statements about cross product in>R^3 are more enlightening
when translated into statements aboutwedge product (or
exterior product). I think your question>is an example. Here
is a rephrasing, with extra padding, in>exterior-algebraic
terms.We are in R^3, with dual space (R^3)* and exterior
algebra>Lambda(R^3), where by construction (or definition)
Lambda(R^3)>is a graded algebra whose degree-0 part is
identified with R,>whose degree-1 part is (R^3)*, and which is
generated as an>exterior algebra by its degree-1 part; and
where, therefore,>by definition (or construction) Lambda(R^3)
is the algebra>of alternating multilinear forms on R^3. Let
lambda be any>covector in R^3, that is, any element of (R^3)*.
Consider>the transformation mu |--> mu ^ lambda from (R^3)*>to
(R^3)* ^ (R^3)* (that is, from the [necessarily
alternating!]>1-forms to the alternating 2-forms; that is,
from the degree-1>part of the exterior algebra to the degree-2
part). Then>this transformation is linear. WHEN ONE USES (for
instance)>A METRIC (or a choice of basis, or ...) TO IDENTIFY
R^3 with>(R^3)*, and consequently to make many other
identifications of>various 3-dimensional vectorspaces, THEN
THE MATRIX OF THISwedge-with-lambda TRANSFORMATION becomes the
matrix you're>interested in.>*gulp*<jill-- To s&e^n]d me
m~a}i]l r%e*m?ov[e bit from my a|d)d:r{e:s]s.===
===
Subject:
: Re:
Is This Right?> For a while, I have been playing with the
automorphism group of the> symmetric group S_6. I believe I
have now derived the sizes of its> conjugacy classes. Does
anyone care to check (here I include S_6 in> Aut(S_6) by
identifying g with conjugation through g)?> 1440=> 1=the
identity> 45=the nontrivial central elements in the
Sylow2-subgroups, which are> the even elements of order 2 in
S_6 (S_6 conjugacy class 2+2+1+1)> 80=the elements of order 3
in S_6 (both 3-cycles and permutations of> S_6-conjugacy class
3+3)> 90=the even elements of order 4 in S_6 (S_6-conjugacy
class 4+2)> 144=the 5-cycles in S_6Everything above here lies
in A_6 =~ PSL(2,9).> 30=the odd elements of order 2 in S_6
(both transpositions and> S_6-conjugacy class 2+2+2)> 90=the
4-cycles in S_6> 240=the elements of order 6 in S_6 (both
6-cycles and S_6-conjugacy> class 3+2+1)And these in S_6A_6.>
144=the elements of order 10 in Aut(S_6)> 36=the elements of
order 2 which arise as 5th powers of elements of> order 10>
180=the elements of order 8 in Aut(S_6) which are 8-cycles in>
Aut(S_6)'s action on 10 pointsAnd these in PGL(2,9)A_6.>
180=the outer automorphisms of order 4> 180=the elements of
order 8 in Aut(S_6) which are of S_10-conjugacy> class 8+2 in
Aut(S_6)'s action on 10 pointsAnd these in M_{10}A_6.> The
addition works out. Hopefully the group theory does also.As
Derek Holt has already noted, you're indeed right. Did you
dothis by hand? If so, I'd be interested to know what
method(s)you used. (Myself, I cheated by looking in _The Atlas
of FiniteGroups_, although I do have a presentation of A_6
buried in mynotes somewhere that should be particularly
amenable to thiskind of investigation.)-- ===
===
Subject:
: Re:
Question about associativity of cartesian product> [ example
of Mathworld-nonsense skipped ]Is Mathworld nonsensical, or
was my (mis?)interpretation of itnonsensical?> As mentioned
above, you will probably find more useful information> in the
library of a nearby university.> Any standard text on category
theory should provide clear definitions, e.g.I gradually became
aware, at first that there was some kind ofpossibly interesting
relationship between categories and the thingswhich I'm trying
to understand, and eventually that they're not justrelated but
in fact the proper context in which to try to understandthose
things. Until a couple weeks ago, category theory was
ameaningless phrase to me and about as interesting as
underwaterChinese basket weaving. My original problem was
trying to understandand formalize polymorphic typing in
computer languages, whicheventually led to me needing to
understand what an equivalencerelation is (don't laugh; Barbie
is right), then to how to formallydefine cartesian product,
then to what a bijection is, then to what anisomorphism is,
and finally to realizing that there even exists aunifying
theme to this stuff, and that category theory is it, so nowI
realize that it's possible to study this stuff without feeling
likeI'm trying to climb onto a floating beach ball. But I had
no way toknow, without first going through this process, that
I needed to studycategory theory. (I've also subsequently
realized that there's alreadymuch published research
specifically on the applications of categorytheory to data
types, meaning that probably I'm not going to have totry to
figure out anything new at all, but just read what
everybodyelse already knows.)> [1] Adamek, Herrlich, Strecker:
Abstract and concrete categories> [2] Freyd, Scedrov:
Categories, Allegories> [3] Mac Lane: Categories for the
working mathematicianThank you for the references; I also
found yesterday ComputationalCategory Theory by Rydeheard and
Burstall which appears to be idealfor me (and available for
free on the 'net too), including formalismof category theory
using a computer language (ML).I don't fully understand the
rest of your message (e.g. I don't knowyet what a functor is),
but I'll reread it after I read at leastComputational Category
Theory. That book also recommends yourreference [3] as a
complement, as well as Herrlich, Strecker:
CategoryTheory.===
===
Subject:
: Re: Question about associativity of
cartesian product>> [ example of Mathworld-nonsense skipped ]>
Is Mathworld nonsensical, or was my (mis?)interpretation of
it> nonsensical?Mathworld was.>[...]> Thank you for the
references; I also found yesterday Computational> Category
Theory by Rydeheard and Burstall which appears to be ideal>
for me (and available for free on the 'net too), including
formalism> of category theory using a computer language (ML).>
I don't fully understand the rest of your message (e.g. I don't
know> yet what a functor is), but I'll reread it after I read
at least> Computational Category Theory. That book also
recommends your> reference [3] as a complement, as well as
Herrlich, Strecker: Category> Theory.If your background is in
CS, the bookBarr, Wells: Category Theory for Computing
Sciencemight interest you as well, see
e.g.<http://www.cwru.edu/artsci/math/wells/pub/ctcs.html===

===
Subject:
: A minor tragedy in 3D non-visualizationTo be able to
visualize 4 Dimensional geometry in more concrete terms,I
wanted at first to be able to see the Invariant of
sphericaltrigonometry D =
Sin[a]/Sin[A]=Sin[b]/Sin[B]=Sin[c]/Sin[C] as somegeometrical
parameter... may be as a dihedral/solid angle, as someratio of
line lengths, areas, volumes or whatever. There is so far
nosatisfactory reply from News groups, and I am beginning to
wonder ifit is already not a minor tragedy in 3D object
comprehension andvisualization from an integrative
perspective. It is not abouttrigonometrical relations binding
the above { 1/D = 2 tan[R]cos[A/2]Cos[B/2]Cos[C/2] } , but
about the geometrical object /property of the spherical
triangle which can only be described, butnot seen.To
illustrate by an example of what I mean,please consider
theexample: Let two secants cut a circle making two intercepts
measuredfrom a pole, (a,A) and (b,B) lengths on each secant. We
have a A = bB = T^2 where T is tangent length from pole to the
circle. If weattempt to seek what a/b or A/B is, firstly it is
not wrong to attempt to do so, nor it is difficult, it is the
similarity ratiobetween the similar triangles having (a,b) and
(A,B) as sidesneighboring the common pole/vertex.In flat
geometries, Sin[a]-> a, Sin[b]-> b , c->Sin[c] etc. and D
iscircum-diameter of the flat triangle. Rightly or wrongly, I
thought itwould to some extent at least break the abstraction
of 4D space whoseintersections and projections are in 3D, that
we could freely gobetween elliptic and hyperbolic geometries on
concrete terms and soon, but I am stuck with 3D visualization
itself.Hope somebody would throw a photon or two on this old
and stillelusive problem of geometric
non-linearity.===
===
Subject:
: Re: A minor tragedy in 3D
non-visualizationsin(A)/a = sin(B)/b = sin(C)/cis the
circumdiameter for a planar trigon? I don't know what you
mean,Sin[a]-> a etc.> To illustrate by an example of what I
mean,please consider the> example: Let two secants cut a
circle making two intercepts measured> from a pole, (a,A) and
(b,B) lengths on each secant. We have a A = b> B = T^2 where T
is tangent length from pole to the circle. If we> attempt to
seek what a/b or A/B is, firstly it is not wrong to > attempt
to do so, nor it is difficult, it is the similarity ratio>
between the similar triangles having (a,b) and (A,B) as sides>
neighboring the common pole/vertex.> In flat geometries,
Sin[a]-> a, Sin[b]-> b , c->Sin[c] etc. and D is>
circum-diameter of the flat triangle. Rightly or wrongly, I
thought
ithttp://www.channel1.com/users/bobwb/synergetics/photos/
x6girdle6.html--Give Earth a Trickier Dick Cheeny -- out of
office, after GIGA
years.http://www.benfranklinbooks.com/http://www.rand.org/
publications/randreview/issues/rr.12.00/http://
members.tripod.com/~american_almanac===
===
Subject:
: Re: A minor
tragedy in 3D non-visualizationThread title of the
month.===
===
Subject:
: Re: Limit of Sequence :
s(n)=(.9)(.99)(.999)...[1-(10^(-n)]> The sequence below
clearly converges. But can we specify the limit?> .9,
(.9)(.99), (.9)(.99)(.999), ...You mean product_{n=1}^infinity
(1-10^{-n}).This is related to Dedekind's eta function (which
comes up in the theoriesof partitions and of modular forms).--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques,
79, 91-92; mistakes his penis for a square root, 88-9Francis
Wheen, _How Mumbo-Jumbo Conquered the World_===
===
Subject:
: how to
obtain simpler form of (1/80)^(1/4) ?I solved a problem and got
a correct answer of (1/80)^(1/4)or in words: the 4th root of
1/80.But both Maple and the book had the answer written in
different ways.The book's answer was: ( (125)^(1/4) ) / 10and
Maple's answer was: (1/10)* ( 5 ^(3/4) )and I am not sure was
simplifications allow one to transform my answer into there's.
In case you're wondering, the initial problem was:(x^4 +
1)^(1/4) = 3xwhich I solved down tox = (1/80)^(1/4)===
===
Subject:
:
Re: integral help>Can someone explain how to solve this
integral analytically:>Int( -tan(t) * sin(t) * dt )THINK a
little!! tan/cos = sin^2/cos = (1-cos^2)/cos = 1/cos - cos=sec
- cos===
===
Subject:
: Re: integral helpEn el
mensaje:405b0bb1.1879513@netnews.worldnet.att.net,mathedman
<mathedman@hotmail.CUT.com> escribi.97:>> Can someone explain
how to solve this integral analytically:>> Int( -tan(t) *
sin(t) * dt )> THINK a little!!> tan/cos = sin^2/cos =
(1-cos^2)/cos = 1/cos - cos> =sec - cosUuum... tan(x)/cos(x) =
sin(x)/cos^2(x)But, that isn`t the OP question ...--
===
===
Subject:
: Re: integral help>Can someone explain how to solve
this integral analytically:>Int( -tan(t) * sin(t) * dt )> THINK
a little!!> tan/cos = sin^2/cos = (1-cos^2)/cos = 1/cos - cos>
=sec - cosFirst of all, why is giving a valid e-mail address
so important to you? Withall the spam, who can blame one for
not posting their actual address to anewsgroup?David
Moran===
===
Subject:
: Re: integral helpEn el
mensaje:xrq6c.37464$Zp.31572@fed1read07,vsgdp
<nospam@nospam.com> escribi.97:> Can someone explain how to
solve this integral analytically:> Int( -tan(t) * sin(t) * dt
)I = Int( -tan(t)*sin(t), t) = -Int(sin^2(t)/cos(t), t) =
-Int(sin^2(t)*cos(t)/cos^2(t), t)Let u = sin(t), du =
cos(t)dt,I = -Int(u^2/(1- u^2), du) = Int(1 + 1/(1 - u^2), du)
= u + Int(1/(1 - u^2), du) = ....-- ===
===
Subject:
: Re: integral
help> Can someone explain how to solve this integral
analytically:> Int( -tan(t) * sin(t) * dt )Int( -tan(t) *
sin(t) * dt ) = = tan(t) * cos(t) - Int(sec^2(t) * cos(t) *
dt) = sin(t) - Int(sec(t) * dt).Jose Carlos Santos===
===
Subject:
:
A rational X irrational gameHelloI've been thinking about the
following problem (or puzzle) but hasn'tcome to a conclusion
yet. Maybe someone can give a hint.A and B are playing the
following game: On the real line, A chooses aclosed interval
I1 of length 0<L1<=1. Then, B chooses a closedinterval I2, of
length 0<L2<=1/2, contained in I1. Then A, in turn,chooses a
closed interval I3, of length 0<L3<=1/3, contained in I2,and
so on. We know there's one, and only one, real number x
thatbelongs to all of the intervals I1, I2, I3.... If x is
rational, Awins the game, and if x is irrational, then B wins.
We are asked tofind an strategy that B should follow so that he
will certainly winthe game no matter how B chooses his
intervals.Since the rationals are countable and the
irrationals are not, itreally seems that there is such a
strategy that assures B will everwin, but I couldn't find it
so far.Thank you.Artur===
===
Subject:
: Re: A rational X irrational
game>A and B are playing the following game: On the real line,
A chooses a>closed interval I1 of length 0<L1<=1. Then, B
chooses a closed>interval I2, of length 0<L2<=1/2, contained
in I1. Then A, in turn,>chooses a closed interval I3, of
length 0<L3<=1/3, contained in I2,>and so on. We know there's
one, and only one, real number x that>belongs to all of the
intervals I1, I2, I3.... If the players agree and cooperate to
converge on a single number, you canpin it down like that. But
they can't be guaranteed to do that (in factthey probably
won't if they are competitive). >If x is rational, A>wins the
game, and if x is irrational, then B wins. We are asked
to>find an strategy that B should follow so that he will
certainly win>the game no matter how B chooses his
intervals.Within any In of length Ln > 0, there will be both
rational and irrationalnumbers. In fact, the cardinality of
the respective sets of remainingrational and irrational
numbers after move n is the same as it was beforemove 1. The
game never ends. >Since the rationals are countable and the
irrationals are not, it>really seems that there is such a
strategy that assures B will ever>win, but I couldn't find it
so far.--Keith Lewis klewis {at} mitre.orgThe above may not
(yet) represent the opinions of my employer.===
===
Subject:
: Re: A
rational X irrational game>>A and B are playing the following
game: On the real line, A chooses a>>closed interval I1 of
length 0<L1<=1. Then, B chooses a closed>>interval I2, of
length 0<L2<=1/2, contained in I1. Then A, in turn,>>chooses a
closed interval I3, of length 0<L3<=1/3, contained in I2,>>and
so on. We know there's one, and only one, real number x
that>>belongs to all of the intervals I1, I2, I3.... > If the
players agree and cooperate to converge on a single number,
you can> pin it down like that. But they can't be guaranteed
to do that (in fact> they probably won't if they are
competitive). How can the final intersection of these nested
intervals, no matterhow they are chosen, ever contain more
than a single number? The nthinterval is shorter than 1/n in
length. If there are two numbers in theintersection, they
differ by some amount, and there is a N so that thereciprocal
1/N is smaller than that difference. That means that fromstep
N on, at most one of those numbers is in the intersection.
Theintersection must be nonempty, since these are closed
intervals of R,and the real line has the property that a
nested sequence of nonemptycompact intervals has nonempty
intersection.>>If x is rational, A>>wins the game, and if x is
irrational, then B wins. We are asked to>>find an strategy that
B should follow so that he will certainly win>>the game no
matter how B chooses his intervals.> Within any In of length
Ln > 0, there will be both rational and irrational> numbers.
In fact, the cardinality of the respective sets of remaining>
rational and irrational numbers after move n is the same as it
was before> move 1. The game never ends. It ends in the limit.
Take 1/2 second for step 1, 1/4 second forstep 2, and so forth
with 1/2^N second for step N. The game is overin one second,
albeit the moves are somewhat rapid towards the end.>>Since
the rationals are countable and the irrationals are not,
it>>really seems that there is such a strategy that assures B
will ever>>win, but I couldn't find it so far.> --Keith Lewis
klewis {at} mitre.org> The above may not (yet) represent the
opinions of my employer.Dale===
===
Subject:
: Re: A rational X
irrational game>Hello>I've been thinking about the following
problem (or puzzle) but hasn't>come to a conclusion yet. Maybe
someone can give a hint.>A and B are playing the following
game: On the real line, A chooses a>closed interval I1 of
length 0<L1<=1. Then, B chooses a closed>interval I2, of
length 0<L2<=1/2, contained in I1. Then A, in turn,>chooses a
closed interval I3, of length 0<L3<=1/3, contained in I2,>and
so on. We know there's one, and only one, real number x
that>belongs to all of the intervals I1, I2, I3.... If x is
rational, A>wins the game, and if x is irrational, then B
wins. We are asked to>find an strategy that B should follow so
that he will certainly win>the game no matter how B chooses his
intervals.>Since the rationals are countable and the
irrationals are not, it>really seems that there is such a
strategy that assures B will ever>win, but I couldn't find it
so far.Hint: B doesn't have to force a win on the first move.
Onthe first move B prevents one way he can lose, on thenext
move he prevents another method of losing...>Thank
you.>Artur===
===
Subject:
: Re: A rational X irrational
game>Hello>I've been thinking about the following problem (or
puzzle) but hasn't>come to a conclusion yet. Maybe someone can
give a hint.>A and B are playing the following game: On the
real line, A chooses a>closed interval I1 of length 0<L1<=1.
Then, B chooses a closed>interval I2, of length 0<L2<=1/2,
contained in I1. Then A, in turn,>chooses a closed interval
I3, of length 0<L3<=1/3, contained in I2,>and so on. We know
there's one, and only one, real number x that>belongs to all
of the intervals I1, I2, I3.... If x is rational, A>wins the
game, and if x is irrational, then B wins. We are asked
to>find an strategy that B should follow so that he will
certainly win>the game no matter how B chooses his
intervals.>Since the rationals are countable and the
irrationals are not, it>really seems that there is such a
strategy that assures B will ever>win, but I couldn't find it
so far.> Hint: B doesn't have to force a win on the first
move. On> the first move B prevents one way he can lose, on
the> next move he prevents another method of losing...Ummm...
based on your hint, this may work:Once the interval I1 has
been choosen by A, B enumerates the rationalsin I1. Let {x_1,
x_2....} be such enumeration. Then, B chooses I2 insuch a way
that x_1 is not in I2 (this is always possible, because
thelength of I2, though positive, is allowed to be arbitrarily
small). Inthe next move, A chooses an I3 contained in I2. And,
in his turn, Bchooses an I4 that doesn't contain x_2 .Carrying
on with this method of choices, B ensures the game willgenerate
a sequence {Im} of closed intervals such that no x_n belongsto
all of the Im's. Therefore, the element x, common to all Im's,
isnot covered by the enumeration {x_1, x_2....}, which implies
x must beirrational. And B wins!Artur===
===
Subject:
: Re: Cantor's
Diagonal Argument
<404ea1b2$0$31901$afc38c87@news.optusnet.com.au
<4en2i1-nuf.ln1@lexi2.athghost7038suus.net
<404ef744$0$3952$afc38c87@news.optusnet.com.au
<40503e3e$0$8359$afc38c87@news.optusnet.com.au
<qf49i1-09k.ln1@lexi2.athghost7038suus.net
<40525513$0$31905$afc38c87@news.optusnet.com.au
<ofnbi1-npl.ln1@lexi2.athghost7038suus.net
<40537499$0$8354$afc38c87@news.optusnet.com.au
<j09ei1-lip.ln1@lexi2.athghost7038suus.net
<4054bb0e.6149272@news.btinternet.com
<4054fa24$0$8357$afc38c87@news.optusnet.com.au
<c2qgi1-5jq.ln1@lexi2.athghost7038suus.net
<40562664$0$15134$afc38c87@news.optusnet.com.au
<8bhii1-mnr.ln1@lexi2.athghost7038suus.net
<4056d6db$0$3957$afc38c87@news.optusnet.com.au
<1s2mi1-i3u.ln1@lexi2.athghost7038suus.net
<4058deb3$0$31901$afc38c87@news.optusnet.com.au
<s0loi1-6sv.ln1@lexi2.athghost7038suus.net
<405a3179$0$3956$afc38c87@news.optusnet.com.auX-SessionID:
TfE6c-7486-45-27568@news.uchicago.eduX-Hash-Info:
post-filter,v:1.4X-Hash: 5e086611 ed4db1e1 5177447b efe1d019
b1559576>Instead of>2 <-> 0.0100000000000000000000...>2 <->
0.0101010101010101010101...by 0.01.. I indicated I mean>
__>0.01..>In that case, I can replace b with the sequence>b =
0.0110111001011101111000100110101011...>where I'm simply
concatenating consecutive integers>into the expansion:>b = 0.0
1 10 11 100 101 110 111 1000 1001 1010 1011...> That's back to
the original problem with irrationals, 1st you> have to
define_the_irrational, my point is definitions are countable,>
since definition = program, numbers are countable. All
irrarationals> are computable yes? To negate this you are
insisting that there> is a massive set of NON COMPUTABLE
IRRATIONALS.First, you are right, you can't define (i.e.,
finitely describe)every irrational. That's simply because
there are more irrationalsthan there are potential
descriptions of irrationals.But second, you are wrong: not
every definition is a program, onlyconstructive definitions
are. In fact, there are irrationals that canbe defined
(finitely described) but which, nevertheless, are
notcomputable. See below.> What does that mean exactly? Can I
see one?Obviously I cannot show you one that cannot be
described. But I candescribe one to you that cannot be
computed: Consider a UniversalTuring Machine M (i.e., a
program taking two natural inputs (i,j) suchthat for every
computable function f from naturals to naturals thereis some i
such that M(f,j) = f(j) for all j). Now consider thebinary
fraction that has a 1 in position i iff M halts on input
(i,i).The so-defined number is not computable.===
===
Subject:
: Re:
Cantor's Diagonal Argument>> Instead of>> 2 <->
0.0100000000000000000000...>> 2 <->
0.0101010101010101010101...> by 0.01.. I indicated I mean>>
__>> 0.01..In that case, I can replace b with the sequenceb =
0.0110111001011101111000100110101011...where I'm simply
concatenating consecutive integers>into the expansion:b = 0.0
1 10 11 100 101 110 111 1000 1001 1010 1011...>That's back to
the original problem with irrationals, 1st you>have to
define_the_irrational, my point is definitions are
countable,>since definition = program, numbers are countable.
All irrarationals>are computable yes? To negate this you are
insisting that there>is a massive set of NON COMPUTABLE
IRRATIONALS.> First, you are right, you can't define (i.e.,
finitely describe)> every irrational. That's simply because
there are more irrationals> than there are potential
descriptions of irrationals.> But second, you are wrong: not
every definition is a program, only> constructive definitions
are. In fact, there are irrationals that can> be defined
(finitely described) but which, nevertheless, are not>
computable. See below.>What does that mean exactly? Can I see
one?> Obviously I cannot show you one that cannot be
described. But I can> describe one to you that cannot be
computed: Consider a Universal> Turing Machine M (i.e., a
program taking two natural inputs (i,j) such> that for every
computable function f from naturals to naturals there> is some
i such that M(f,j) = f(j) for all j). Now consider the> binary
fraction that has a 1 in position i iff M halts on input
(i,i).> The so-defined number is not computable.I was an
advocate of Cantor for a decade. I'm discussing the above
'number'in reply to George's post, basically from the POV
Halt(x) doesn't exist anyway.We could construct a similar
number.x1 = 1 iff God is real, zero otherwise. (non
computable!)x2 = 1 iff there are 2 gods of equal powerx3 = 1
iff there are 3 gods of equal powerI added the ...of equal
power so more than one proposition can hold.Now we can
construct the non computable irrational!longlastingduels =
E(n) 2^-xnIs that a number?Herc===
===
Subject:
: Re: Cantor's
Diagonal Argument>That's back to the original problem with
irrationals, 1st you>have to define_the_irrational, my point
is definitions are countable,>since definition = program,
numbers are countable. All irrarationals>are computable yes?
To negate this you are insisting that there>is a massive set
of NON COMPUTABLE IRRATIONALS.>What does that mean exactly?
Can I see one?> No you can't, that's the point. Most real
numbers are indescribable> irrational numbers. They have
infinitely many digits, and cannot be> computed by a
finite-sized program or described in a finite-sized Usenet>
message.But my list of computable numbers is infinite, I can
describe some of them.Infinite length is not a problem, just
specifiy what you can.Herc===
===
Subject:
: Re: Cantor's Diagonal
Argument <c3dfvo$m11$1@newslocal.mitre.org
<405a6074$0$25292$afc38c87@news.optusnet.com.au>That's back to
the original problem with irrationals, 1st you>have to
define_the_irrational, my point is definitions are
countable,>since definition = program, numbers are countable.
All irrarationals>are computable yes? To negate this you are
insisting that there>is a massive set of NON COMPUTABLE
IRRATIONALS.>>What does that mean exactly? Can I see one?>>
No you can't, that's the point. Most real numbers are
indescribable>> irrational numbers. They have infinitely many
digits, and cannot be>> computed by a finite-sized program or
described in a finite-sized Usenet>> message.>But my list of
computable numbers is infinite, I can describe some of them.If
you can specify how to compute a number, that counts as a
description. I'm talking about the ones which are impossible
to describe in any way. Thedigits don't repeat or fit any
other kind of pattern, they aren't the squareroot of
something, not related to pi. These numbers have no
practicalpurpose but they're there in between the numbers that
we can describe.>Infinite length is not a problem, just
specifiy what you can.But Cantor's theorem when applied using
base 10 proves that for every numberyou add to the list, the
ratio of numbers missed to numbers on the listgrows by a
factor of 8. You can also apply it using base 16 and the
ratiowill grow by a factor of 14 each time through. --Keith
Lewis klewis {at} mitre.orgThe above may not (yet) represent
the opinions of my employer.===
===
Subject:
: Re: Cantor's Diagonal
Argument> : That's back to the original problem with
irrationals, 1st you> : have to define_the_irrational> I'm
sorry, you just CAN'T SAY that.> Seriously, in that sentence,
define is hopelessly ambiguous.> You have to invent 2 new
different words for the two meanings> BETWIXT which define is
ambiguous, and then REPLACE define> with whichever one of them
is appropriate, wherever it occurs,> in order to talk
coherently about this.> I mean, I can DEFINE the irrational,
pi asthe ratio of the circumference of a circle to its>
diameter, but that just plain ISN'T HELPING anybody.> I can
define it as the output of a certain Turing Machine> but even
that is a little dodgy since that machine would> have to run
forever in order for its output to be the> actual value of pi
as opposed to some finite approximation thereof.> In order to
clarify the difference between the two meanings of define,>
let's eliminate confusions caused by infinity from the
scenario.> Let's just pick some random expression that refers
to a number> that YOU DON'T KNOW, such as the number of
Jefferson's> descents. That is just as good a definition of
227 as227 is, if you KNOW that Jefferson has 227 descents.>
But if you DON'T KNOW that, then this ISN'T a good definition
of> 227. The phrase I chose INDICATES 227: it MEANS 227; it
REFERS> to 227, whether you know that or not. But it doesn't
EXPLICATE> 227; it doesn't REPRESENT 227; it doesn't SPECIFY
227.227, on the other hand, if you know what number-base
you're> in, and you know we're using the place-value paradigm
for> expressing natural numbers, is as full an explication of>
this number as you could ask.> Now, on one level, you could say
that Pi could be indicated> (since the finite TM program for
the TM that outputs it> is finitely describable and
understandable), but that it> can't be represented or
specified because its specification> in
numerical-terms-we-can-use is infinite, and therefore
pragmatically> UNuseable. In real life, however, we won't be
that uncharitable;> we will allow the finite specification of
the TM program to COUNT> as a specification of Pi. Some
irrationals, though, don't have> ANY TM program that will
specify them, not EVEN when it runs forever.> What you are
asking is for us to INDICATE one of those irrationals that> we
*can't* EXPLICATE, NOT EVEN via some lame dodge like calling it
the> infinite output of a finite TM.> Here is 1 such
indication:> The irrational in [0,1] represented by the
bit-string> that has a 1 in its nth place whenever n encodes a
TM that> doesn't halt when its input is n, and has an 0 in all
other places.> In other words, DON'T SAY define when you are
trying to> discuss this stuff. Say point at for a descriptive
definition> and say specify for a representative one.Ok, so
Cantor illustrated a non defined number that's not computable.
Thesenon terminating, non repeating, non patterned sequences
force us to accepta higher class than infinity, an infinity
higher of infinities.> Here is 1 such indication:> The
irrational in [0,1] represented by the bit-string> that has a
1 in its nth place whenever n encodes a TM that> doesn't halt
when its input is n, and has an 0 in all other places.It crops
up every time its like the 'witness' to uncountable. Is thisthe
only specified_irrational number that supports Cantors
argument?Its like this statement has no proof forces us to
discard formal mathematics.Halt does not exist. There is only
an oracle that gives its values.The Halting proof begins with
Halt having 2 values : 0 or 1.The proof then shows it doesn't
exist, some values can be manually populated.In reality it has
2 known values plus an undetermined_as_yet.We could construct a
similar number.x1 = 1 iff God is real, zero otherwise. (non
computable!)x2 = 1 iff there are 2 gods of equal powerx3 = 1
iff there are 3 gods of equal powerI added the ...of equal
power so more than one proposition can hold.Now we can
construct the non computable irrational!longlastingduels =
E(n) 2^-xnIs that a number?Herc===
===
Subject:
: Re: Cantor's
Diagonal Argument : > What you are asking is for us to
INDICATE one of those irrationals that : > we *can't*
EXPLICATE, NOT EVEN via some lame dodge like calling it the :
> infinite output of a finite TM. : : > Here is 1 such
indication: : : > The irrational in [0,1] represented by the
bit-string : > that has a 1 in its nth place whenever n
encodes a TM that : > doesn't halt when its input is n, and
has an 0 in all other places. : : : > In other words, DON'T
SAY define when you are trying to : > discuss this stuff. Say
point at for a descriptive definition : > and say specify for
a representative one. : Ok, so Cantor illustratedNo, do NOT
say illustrate. I WARNED you that natural langaugeis
worthlessly ambiguous in this case. If we are going to talk
thenwe are going to have to AGREE to SHARE words for concepts
that areNOT handled right in the usual dictionary. Cantor
successfullyPOINTED AT : a non defined numberBull! I TOLD you
NOT TO SAY *defined*!By the usual DICTIONARY definition of
defined,Cantor's pointing (and MY indication above) IS a
definition! : that's not computable.Right. While whether this
number is definable is debatableand contested, whether it's
computable isn't: it definitely isn't. : These non
terminating, non repeating, non patterned sequences force us
to accept : a higher class than infinity, an infinity higher
of infinities.No, just 1 will do. : > Here is 1 such
indication: : : > The irrational in [0,1] represented by the
bit-string : > that has a 1 in its nth place whenever n
encodes a TM that : > doesn't halt when its input is n, and
has an 0 in all other places. : : It crops up every time its
like the 'witness' to uncountable.Not every time; every
formulation is different; this isn't whatcropped up in
Cantor's time. : Is this : the only specified_irrational
number that supports Cantors argument?Use the words RIGHT,
dammit! This number CANNOT BE *specified*! That's THE WHOLE
POINT! And no, of course it's not the only one. What youwere
trying to ask is, is it the only one that can be pointed at,
thatcan be meaningfully indicated in natural language. And,
again, theanswer is, of course not. : Its like this statement
has no proof forces us to discard formal mathematics.We don't
discard it. We just acknowledge that classical first-order
logichas limitations. This doesn't force us to discard
anything that lies WITHINthose limits. : Halt does not
exist.That is not only idiotic, it is ungrammatical.Halt IS A
VERB. It therefore cannot, unquoted, bethe subject of a
sentence in English. Lots of TM's halt,so, OBVIOUSLY, Halt
exists. : There is only an oracle that gives its values.No,
there isn't.Gee, that was easy.In fact, there is no such thing
as an oracle.That was even easier. : The Halting proof begins
with Halt having 2 values : 0 or 1.Why are you capitalizing
Halt, like it was a noun? Haltis a VERB! : The proof then
shows it doesn't exist,No, the proof shows that A TM THAT
TELLS WHETHER ITS INPUTIS OR ISN'T A TM,input WHERE THE TM
HALTS ON THE INPUTdoesn't exist. THAT, whatever it is, is NOT
Halt.There is NO SUCH THING as Halt. Halt IS A VERB.It is not
something you can thingify. A TM that tells whethersomething
halts, YES, THAT is a noun, that can either exist or not.And
it doesn't. : some values can be manually populated.No, they
can't. That is NOT what the failure-of-a-TM-to-exist means. :
In reality it has 2 known values plus an
undetermined_as_yet.No, I'm sorry, every TM, on every input,
either halts or itdoesn't. The fact that you don't yet know
which DOES NOT CHANGEthat fact.-- --- The history of our
nation has demonstrated that separate is seldom, if ever,
equal.===
===
Subject:
: Re: Cantor's Diagonal Argument> : Is this> :
the only specified_irrational number that supports Cantors
argument?> Use the words RIGHT, dammit! This number CANNOT BE
*specified*! That'sSpecify is not to define, or to declare
that the specification is computable.If Cantor did not specify
any numbers at all then he has no case. I willuse the term
specify as the weak form of define you where asking for. In
practicespecification is made by AND/OR with the client, the
person presenting the case,its from the specification that the
programmer attempts_to_define the objective.A specification is
just a set of constraints, meaningful or not, nothing more.Its
step 2 in the 8 step software development cycle.1 Scope2
Specification3 Program4 Test5 Document6 Hand Over7 Maintain8
FinaliseThink of specify as point_to, I'm sure the natural
language meanings tie these 2.I'm not referring to formal
specification, which is not a complete science as thespecs can
be compiled to programs and are effectively a redunt phase.Your
halting_number is specified is it not?> THE WHOLE POINT! And
no, of course it's not the only one. What you> were trying to
ask is, is it the only one that can be pointed at, that> can
be meaningfully indicated in natural language. And, again,
the> answer is, of course not.Others?> : Its like this
statement has no proof forces us to discard formal
mathematics.> We don't discard it. We just acknowledge that
classical first-order logic> has limitations. This doesn't
force us to discard anything that lies WITHIN> those limits.>
: Halt does not exist.> That is not only idiotic, it is
ungrammatical.Halt IS A VERB. It therefore cannot, unquoted,
be> the subject of a sentence in English. Lots of TM's halt,>
so, OBVIOUSLY, Halt exists.Its the most fundamental computer
theoretical *function* known, in futureI will write Halt() for
a non existent function taking 2 parameters,Halt(programx,
argumenty).> : There is only an oracle that gives its values.>
No, there isn't.> Gee, that was easy.An Oracle exists is a self
evident statement used in complexity theory.It means
there_isn't as we know it.> In fact, there is no such thing as
an oracle.> That was even easier.prove it if you are so
inclined> : The Halting proof begins with Halt having 2 values
: 0 or 1.> Why are you capitalizing Halt, like it was a noun?
Halt> is a VERB!> : The proof then shows it doesn't exist,>
No, the proof shows that A TM THAT TELLS WHETHER ITS INPUT> IS
OR ISN'T A TM,input WHERE THE TM HALTS ON THE INPUT> doesn't
exist. THAT, whatever it is, is NOT Halt.> There is NO SUCH
THING as Halt. Halt IS A VERB.> It is not something you can
thingify. A TM that tells whether> something halts, YES, THAT
is a noun, that can either exist or not.> And it doesn't.well
1000s of textbooks call that TM Halt(). No one says, the TM
thattells whether its input is or isn't a TM and where the TM
halts on its inputdoesn't exist. They define the function,
give it a name, let it be modifiedand referred_to by a second
function, then conclude.The proof begins :Define Halt(x, y)
having outputs 0 and 1.A contradiction is then found.Get it?
the outputs 0 and 1 didn't work! Thats the part that got
disproven.> : some values can be manually populated.> No, they
can't. That is NOT what the failure-of-a-TM-to-exist means.But
you are using the output of the nonexistent TM to form a
number.IN FACT, whether TM(1) halts or not MAY be
determined.Whether TM(2) halts or not MAY be determined.We can
examine each program indepently and calculate *some* of
thevalues of your halting_number. Sometimes we can notice a TM
halt, sometimeswe can prove it will eventually halt, sometimes
we can prove it will never halt.We just can't (systematically)
get a 0 or 1 value for EVERY program.Halt( TMx, y ) = 0 means
TMx(y) will go into an infinite loop.I could easily populate
the values of the early indexed TMs.Halt(1,1) = 1Halt(1,2) =
1Halt(1,3) = 1why ? because one state machines always land on
the finish state.Halt(2,1) = 1Halt(2,2) = 1These are correct
so far aren't they? They are the initial values of you
halting_numberaren't they? We can populate SOME of the values
can't we? Does the haltingproof insist we know NOTHING about
program behavour?> : In reality it has 2 known values plus an
undetermined_as_yet.> No, I'm sorry, every TM, on every input,
either halts or it> doesn't. The fact that you don't yet know
which DOES NOT CHANGE> that fact.In reality we work on what we
KNOW. The array in reality can be populatedwith some KNOWN
values. To do this IN PRACTICE would require aternary
output.Programs are complex, basically the Halting proof picks
up on the fact thata program can feasibly generate its own
functions on the fly that aren't amenibleto inspection without
running the code. Some of the values are not known,they will
never be known, just a dozen state TMs *would* run for the
life of the universeuntil they halt.So why is this
impossible_to_populate every given point array any different
to this array?x1 = 1 iff God is real, zero otherwise. (non
computable!)x2 = 1 iff there are 2 gods of equal powerx3 = 1
iff there are 3 gods of equal powerlonglastingduels = E(n)
2^-xnHalt(1039383784748) = unknown, either 0 or
1x1039383784748 = unknown, either 0 or 1The fact is we don't
know things, we can construct locii from such parameters, not
numbers.Herc===
===
Subject:
: Re: Interpolating Function>If I know
my function f has this property:> f(0) = a(0)> f'(0) = a(1)>
f''(0) = a(2)> ....>where a(0), a(1), ... are parameters>I
wonder whether there is any method so that I can derive my
function f.>Does anyone know or have any ideas on this? Thank
you!> f(x) = the sum, from k = 0 to whatever, of a(k) times
x-to-the-k,> divided by k-factorial.> Gerry Myerson
(gerry@maths.mq.edi.ai) (i -> u for email) Keeping one's
fingers crossed, of course. There is always an infinite setof
solutions. (Recall the monster function exp(-1/x^2), defined
as 0 at 0. It has allthe derivatives equal to 0 at 0, and is
non-zero away from x=0.) (1) If the number of conditions is
finite, the sum described there givesa polynomial of the
lowest degree fitting the conditions. (2) If the number of
conditions is infinite, the series may not convergeaway from
x=0 (take a(n)=n!, the factorial), but there will be a
functionsatisfying the conditions, obtained by a tricky
construction.Cheers, ZVK(Slavek).===
===
Subject:
: Re: Interpolating
FunctionI'm really sorry for my crappy question. Indeed, I was
thinking about theTaylor expansion that Gerry suggested but I
want to convert it to a reducedform (not infinite series)
provided some properties of a(n). For example: ifa(n) = 1/n!
then f(x) = e^xI think this is used a lot in probability. For
my specific project:a(0), a(1),.... are the distribution of a
variable. We also know theexpected value, which is
sigma(k*a(k)) = e and we want to estimatesigma(a(k)*a^k) for a
known constant a.If you suggest any good books talking about
generating functions like this,I'd appreciate too.===
===
Subject:
:
continuous mapsThe following statements are know to be
equivalent: f continuous for all A, cl f^-1(A) subset f^-1(cl
A) (a) for all A, f^-1(int A) subset int f^-1(A) (b)Are the
following statements: for all A, f(cl A) subset cl f(A) (1)
for all A, int f(A) subset f(int A) (2)also equivalent to f is
continuous?-- proof: f continuous iff (1)Starting with for all
A, cl f^-1(A) subset f^-1(cl A) for all B, cl f^-1(f(B))
subset f^-1(cl f(B))f(cl B) subset f(cl f^-1(f(B))) subset
ff^-1(cl f(B)) subset cl f(B)Starting with for all A, f(cl A)
subset cl f(A) for all B, f(cl f^-1(B)) subset cl ff^-1(B)cl
f^-1(B) subset f^-1f(cl f^-1(B)) subset f^-1(cl ff^-1(B))
subset f^-1(cl f(B)-- remaining question: f continuous iff
(2)The above method of proof for f continuous iff (1) fails to
establish f continuous iff (2)which was unexpected as (1) & (2)
are dual statements.Any explaination, counter example or
proof?----===
===
Subject:
: Re: continuous mapsWilliam Elliot a
.8ecrit:> The following statements are know to be equivalent:>
f continuous> for all A, cl f^-1(A) subset f^-1(cl A) (a)> for
all A, f^-1(int A) subset int f^-1(A) (b)> Are the following
statements:> for all A, f(cl A) subset cl f(A) (1)> for all A,
int f(A) subset f(int A) (2)> also equivalent to f is
continuous?> -- proof: f continuous iff (1)> Starting with>
for all A, cl f^-1(A) subset f^-1(cl A)> for all B, cl
f^-1(f(B)) subset f^-1(cl f(B))> f(cl B) subset f(cl
f^-1(f(B))) subset ff^-1(cl f(B)) subset cl f(B)> Starting
with> for all A, f(cl A) subset cl f(A)> for all B, f(cl
f^-1(B)) subset cl ff^-1(B)> cl f^-1(B) subset f^-1f(cl
f^-1(B)) subset f^-1(cl ff^-1(B))> subset f^-1(cl f(B)> --
remaining question: f continuous iff (2)> The above method of
proof for f continuous iff (1)> fails to establish f
continuous iff (2)> which was unexpected as (1) & (2) are dual
statements.> Any explaination, counter example or proof?>
----It's false.Let f=Id : E,T -> E,Dwhere T is trivial
topology (empty set and E are the only open sets)and D :
discrete topology (everey subset of E is open)If is an open
map but not continuous.open map : the direct image of an open
set is an open set (equivalent to your statement)If all your
equivalence were true, the inverse function of a continuous
function would be continuous wich is a well kown wrong
statement.===
===
Subject:
: Re: continuous maps>The following
statements are know to be equivalent:> f continuous> for all
A, cl f^-1(A) subset f^-1(cl A) (a)> for all A, f^-1(int A)
subset int f^-1(A) (b)>Are the following statements:> for all
A, f(cl A) subset cl f(A) (1)> for all A, int f(A) subset
f(int A) (2)>also equivalent to f is continuous?>-- proof: f
continuous iff (1)>Starting with> for all A, cl f^-1(A) subset
f^-1(cl A)> for all B, cl f^-1(f(B)) subset f^-1(cl f(B))>f(cl
B) subset f(cl f^-1(f(B))) subset ff^-1(cl f(B)) subset cl
f(B)>Starting with> for all A, f(cl A) subset cl f(A)> for all
B, f(cl f^-1(B)) subset cl ff^-1(B)>cl f^-1(B) subset f^-1f(cl
f^-1(B)) subset f^-1(cl ff^-1(B))> subset f^-1(cl f(B)>--
remaining question: f continuous iff (2)>The above method of
proof for f continuous iff (1)> fails to establish f
continuous iff (2)>which was unexpected as (1) & (2) are dual
statements.>Any explaination, counter example or proof?>---->
It's false.> Let f=Id : E,T -> E,D> where T is trivial
topology (empty set and E are the only open sets)> and D :
discrete topology (everey subset of E is open)That's not a
counterexample: f is not continuous, nor does it satisfy (2).
Suppose both spaces are R in the usual topology and f(x) = 0,
x <= 0, f(x) = 1, x > 0. Then f is not continuous, but it's
range is two points so it satisfies (2) trivially.===
===
Subject:
:
Re: continuous mapsIn the other direction, let f : R^2 -> R be
the continuous map given by f(x,y) = x. Set A = R x {0}. Then
f(A) = R, so int(f(A)) = R. But A has empty interior, so (2)
fails.===
===
Subject:
: Re: how to obtain simpler form of
(1/80)^(1/4) ?> I solved a problem and got a correct answer
of> (1/80)^(1/4)> or in words: the 4th root of 1/80.> But both
Maple and the book had the answer written in different ways.>
The book's answer was:> ( (125)^(1/4) ) / 10> and Maple's
answer was:> (1/10)* ( 5 ^(3/4) )> and I am not sure was
simplifications allow one to transform my answer > into
there's.80 = 2^4 * 580^(1/4) = 2 * 5^(1/4)(1/80)^(1/4) = 1 /
80^(1/4) = 1 / [2 * 5^(1/4)] = 5^(3/4) / {[2 * 5^(1/4)] *
5^(3/4)}The last step is called rationalizing the
denominator.-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu===
===
Subject:
: Re: how to obtain
simpler form of (1/80)^(1/4) ?> I solved a problem and got a
correct answer of> (1/80)^(1/4)> or in words: the 4th root of
1/80.> But both Maple and the book had the answer written in
different ways.> The book's answer was:> ( (125)^(1/4) ) / 10
= ( (125)^(1/4) ) / ((10^4)^(1/4)) = ( (125/(10^4))^(1/4)> and
Maple's answer was:> (1/10)* ( 5 ^(3/4) )I assume you can see
how Maple's answer is the same as the book's.===
===
Subject:
:
Particular solution to differential equationI am a senior
high-school. In this year's mathematics curriculum, we
areintroduced to differential equations in the form ofy' = ay
+bI know how the general solution is, and how to find it.
However, on ourexams, we get to solve equations like this
one:y' - y=2( e^(2x)-1 ) (E)Since we are not taught how to
solve this directly, we also are given aparticular solution of
the equation (E), in this case:h(x)=2xe^(2x) +1Thus we first
need to prove that h is indeed a solution of (E), then
provethat y = z + h _if and only if_ z is a solution to
z'-z=0. At that point,finding z and deducing y is easy.I know
perfectly well how to deal with such exercises, and
usuallydon't do any mistakes. However I am wondering how it
would be possible tofind that particular solution, h(x). Do
you have any information about it?How should I search the
web?BB===
===
Subject:
: Re: Particular solution to differential
equation>I am a senior high-school. In this year's mathematics
curriculum, we are>introduced to differential equations in the
form of>y' = ay +b>I know how the general solution is, and how
to find it. However, on our>exams, we get to solve equations
like this one:>y' - y=2( e^(2x)-1 ) (E)>Since we are not
taught how to solve this directly, we also are given
a>particular solution of the equation (E), in this
case:>h(x)=2xe^(2x) +1>Thus we first need to prove that h is
indeed a solution of (E), then prove>that y = z + h _if and
only if_ z is a solution to z'-z=0. At that point,>finding z
and deducing y is easy.>I know perfectly well how to deal with
such exercises, and usually>don't do any mistakes. However I am
wondering how it would be possible to>find that particular
solution, h(x). Do you have any information about it?>How
should I search the web?>BB Have you been taught how to
solve(*) y' + f(x)y = g(x) ?The left side of (*) is linear in
y and y'. The right side depends only on x.Look up integrating
factor or linearin the index to your textbook. The trick to
solving (*) is to findan antiderivative F(x) of f(x). That is,
findF(x) with F'(x) = f(x). Multiply both sidesof (*) by
exp(F(x)): exp(F(x))y' + exp(F(x))f(x)y = exp(F(x))g(x)The
left side is the derivative of exp(F(x))y(use the product rule
and chain rule).You can integrate both sides with respect to x.
For your application f(x) = -1 (constant)and g(x) = 2*(exp(2*x)
- 1). One choice of F(x)is 10-x. Multiply by exp(10-x):
exp(10-x)y' - exp(10-x)y = 2*exp(10-x)*(exp(2*x) - 1) =
2*exp(10+x) - 2*exp(10-x)Integrate both sides wrt x exp(10-x)y
= 2*exp(10+x) + 2*exp(10-x) + CSolve for y y = 2*exp(2*x) + 2 +
C*exp(x-10) Your particular solution is not of this form. I
suspect you intended to solve y' - 2*y = 2*(exp(2*x) - 1).I'll
let you solve this yourself.-- John Adams served two terms as
Vice President and one as President, but lostreelection. Later
his son became President despite losing the popular vote.That
son lost his reelection attempt badly. Now history is
repeating itself.pmontgom@cwi.nl Microsoft Research and CWI
Home: San Rafael, California===
===
Subject:
: Re: Particular
solution to differential equation> I am a senior high-school.
In this year's mathematics curriculum, we are> introduced to
differential equations in the form of> y' = ay +b> I know how
the general solution is, and how to find it. However, on our>
exams, we get to solve equations like this one:> y' - y=2(
e^(2x)-1 ) (E)> Since we are not taught how to solve this
directly, we also are given a> particular solution of the
equation (E), in this case:> h(x)=2xe^(2x) +1> Thus we first
need to prove that h is indeed a solution of (E), then prove>
that y = z + h _if and only if_ z is a solution to z'-z=0. At
that point,> finding z and deducing y is easy.> I know
perfectly well how to deal with such exercises, and usually>
don't do any mistakes. However I am wondering how it would be
possible to> find that particular solution, h(x). Do you have
any information about it?> How should I search the
web?Research is often a useful idea. Two common methods are
undeterminedcoefficients and variation of parameters.P.S.
Variation of parameters is used even on other planets!
http://www.math.ohio-state.edu/~edgar/movie/DayTheEarth.html--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/===
===
Subject:
:
Gaussian-Jordon Reduction C or C++ LibraryI am looking for a C
or C++ library that supports Linear Systems ofequations. I have
a real-time application that needs to reduce 10's ofthousands
of simultaneous equations quickly. I am looking for
speedoptimized reduction algorithms like Gaussian-Jordon
elimination or reductiontechniques encapsulated in a library.
Can you please assist me in findingthis library or more
appropriate news group.===
===
Subject:
: Re: simple integration
question (...it's been a while) Adjunct Assistant Professor at
the University of Montana.>You could let >u = n^x>Then,>ln u =
x ln n>So,>du / u = ln n dx. Which means that [ln n) / u] du =
dxCareful. That should be [1/u*ln(n)] = dx... no wonder my
solutiondiffered from yours...Phew. Two people came up with
n^x*ln(n) + C, and I was starting towonder if my mind was
gone... again...>Now, substituting into the integral
yield>Int( n^x dx) = Int( u [ln n) / u] du )>= Int( ln n du) =
u ln n + CSo this should be Int (du/ln(n)) = u/ln(n) + C.--
==============================================================
========It's not denial. I'm just very selective about what I
accept as reality. --- Calvin (Calvin and
Hobbes)=======================================================
==================
===
Subject:
: seeking a reference regarding
matrix eigenvectorstheory: the eigenvectors associated with
distinct eigenvalues of a Hermitianmatrix are orthogonal. I
browsed the well-known book titled Matrix Analysis byHorn and
Johnson and couldn't locate. But I'm not really familiar with
thisbook. So maybe I just missed it. Can anyone help me find
this theorem?Wenyan===
===
Subject:
: Re: Unknown Functions &
Einstein's Incompetence> Unknown Functions & Einstein's
Incompetence (FAQ)> (c) Eleaticus/Oren C. Webster[snip trolled
garbage]Originally trolled across sci.physics
sci.physics.relativityalt.physics sci.math sci.answers
alt.answers
news.answershttp://b5.sdvc.uwyo.edu/bab5/snds/
argcstpd.wavPsychotic ineducable boring troll Eleaticus, You
see yourself this
way,http://www.mazepath.com/uncleal/effete6.jpg The entire
remainder of the planet sees you this
way,http://www.mazepath.com/uncleal/effete3.jpghttp://
w0rli.home.att.net/youare.swfhttp://www.mazepath.com/uncleal/
sunshine.jpghttp://www.you-moron.com/http://www.apa.org/
journals/psp/psp7761121.htmlhttp://insti.physics.sunysb.edu/~
siegel/quack.html<http://www.firehead.org/~jessh/film/kubrick/
Kubrick-Psycho.html<http://www.naturalchild.com/elliott_barker
/prisons.htmlHey, stooopid troll Eleaticus - Do you want
EVIDENCE? Each of the 24GPS satellites carries either four
cesium atomic clocks or threerubidum atomic clocks in orbit,
with full relativistic correctionsbeing
applied.http://arXiv.org/abs/hep-th/0307140 GR structure,
especially Part 4/p.
7<http://rattler.cameron.edu/EMIS/journals/LRG/Articles/
Volume4/2001-4will/index.html Experimental constraints on
General
Relativity.<http://tycho.usno.navy.mil/ptti/ptti2002/paper20.
pdfhttp://www.eftaylor.com/pub/projecta.pdf Relativity in the
GPS systemhttp://arXiv.org/abs/gr-qc/9909014 falling
lighthttp://www.hawaii.edu/suremath/SRtwinParadox.html<http://
physics.syr.edu/courses/modules/LIGHTCONE/twins.html Twin
Paradoxhttp://arXiv.org/abs/astro-ph/0401086http://arxiv.org/
abs/astro-ph/0312071 Deeply relativistic neutron star
binarieshttp://arXiv.org/abs/gr-qc/0301024 Nordtvedt EffectNIM
A 355 537 (1995)Physics Letters B 328 103 (1994)Physical Review
Letters 64 1697 (1990)Physical Review Letters 39 1051
(1977)Physical Review 135 B1071 (1964)Physics Letters 12 260
(1964)Europhysics Letters 56(2) 170-174 (2001)General
Relativity and Gravitation 34(9) 1371
(2002)http://fourmilab.to/etexts/einstein/specrel/specrel.pdf<
http://www.geocities.com/physics_world/sr/ae_1905_error.htm<
http://www.physics.gatech.edu/people/faculty/finkelstein/
relativity.pdfhttp://users.powernet.co.uk/bearsoft/Paper6.
pdfhttp://users.powernet.co.uk/bearsoft/LPHrel.html
Longitudinal and transverse
masshttp://www.navcen.uscg.gov/pubs/gps/gpsuser/
gpsuser.pdfhttp://www.navcen.uscg.gov/pubs/gps/sigspec/
default.htmhttp://www.navcen.uscg.gov/pubs/gps/icd200/
default.htmhttp://www.trimble.com/gps/index.htmlhttp://
sirius.chinalake.navy.mil/satpred/http://www.phys.lsu.edu/mog/
mog9/node9.htmlhttp://egtphysics.net/GPS/RelGPS.htmhttp://
www.schriever.af.mil/gps/Current/current.oa1http://
edu-observatory.org/gps/gps_books.html<http://
www-astronomy.mps.ohio-state.edu/~pogge/Ast162/Unit5/gps.html-
-Uncle
Alhttp://www.mazepath.com/uncleal/qz.pdfhttp://
www.mazepath.com/uncleal/eotvos.htm (Do something naughty to
physics)===
===
Subject:
: Re: Einstein (1905) Absurdities> Einstein
(1905) Absurdities> (c) Eleaticus/Oren C. Webster>
Thnktank@concentric.net[snip 1300 lines of trolled
garbage]Originally trolled across sci.physics
sci.physics.relativityalt.physics sci.math sci.answers
alt.answers
news.answershttp://b5.sdvc.uwyo.edu/bab5/snds/
argcstpd.wavPsychotic ineducable boring troll Eleaticus, You
see yourself this
way,http://www.mazepath.com/uncleal/effete6.jpg The entire
remainder of the planet sees you this
way,http://www.mazepath.com/uncleal/effete3.pnghttp://
w0rli.home.att.net/youare.swfhttp://www.mazepath.com/uncleal/
sunshine.jpghttp://www.you-moron.com/http://www.apa.org/
journals/psp/psp7761121.htmlhttp://insti.physics.sunysb.edu/~
siegel/quack.html<http://www.firehead.org/~jessh/film/kubrick/
Kubrick-Psycho.html<http://www.naturalchild.com/elliott_barker
/prisons.htmlHey, stooopid troll Eleaticus - Do you want
EVIDENCE? Each of the 24GPS satellites carries either four
cesium atomic clocks or threerubidum atomic clocks in orbit,
with full relativistic correctionsbeing
applied.http://arXiv.org/abs/hep-th/0307140 GR structure,
especially Part 4/p.
7<http://rattler.cameron.edu/EMIS/journals/LRG/Articles/
Volume4/2001-4will/index.html Experimental constraints on
General
Relativity.<http://tycho.usno.navy.mil/ptti/ptti2002/paper20.
pdfhttp://www.eftaylor.com/pub/projecta.pdf Relativity in the
GPS systemhttp://arXiv.org/abs/gr-qc/9909014 falling
lighthttp://www.hawaii.edu/suremath/SRtwinParadox.html<http://
physics.syr.edu/courses/modules/LIGHTCONE/twins.html Twin
Paradoxhttp://arXiv.org/abs/astro-ph/0401086http://arxiv.org/
abs/astro-ph/0312071 Deeply relativistic neutron star
binarieshttp://arXiv.org/abs/gr-qc/0301024 Nordtvedt EffectNIM
A 355 537 (1995)Physics Letters B 328 103 (1994)Physical Review
Letters 64 1697 (1990)Physical Review Letters 39 1051
(1977)Physical Review 135 B1071 (1964)Physics Letters 12 260
(1964)Europhysics Letters 56(2) 170-174 (2001)General
Relativity and Gravitation 34(9) 1371
(2002)http://fourmilab.to/etexts/einstein/specrel/specrel.pdf<
http://www.geocities.com/physics_world/sr/ae_1905_error.htm<
http://www.physics.gatech.edu/people/faculty/finkelstein/
relativity.pdfhttp://users.powernet.co.uk/bearsoft/Paper6.
pdfhttp://users.powernet.co.uk/bearsoft/LPHrel.html
Longitudinal and transverse
masshttp://www.navcen.uscg.gov/pubs/gps/gpsuser/
gpsuser.pdfhttp://www.navcen.uscg.gov/pubs/gps/sigspec/
default.htmhttp://www.navcen.uscg.gov/pubs/gps/icd200/
default.htmhttp://www.trimble.com/gps/index.htmlhttp://
sirius.chinalake.navy.mil/satpred/http://www.phys.lsu.edu/mog/
mog9/node9.htmlhttp://egtphysics.net/GPS/RelGPS.htmhttp://
www.schriever.af.mil/gps/Current/current.oa1http://
edu-observatory.org/gps/gps_books.html<http://
www-astronomy.mps.ohio-state.edu/~pogge/Ast162/Unit5/gps.html-
-Uncle
Alhttp://www.mazepath.com/uncleal/qz.pdfhttp://
www.mazepath.com/uncleal/eotvos.htm (Do something naughty to
physics)===
===
Subject:
: Integration questionHow do i integrate
x^(-1/2)exp^(-x) with respect to x and x goes from 0
toinfinity ? The answer is supposed to be Sqrt[Pi], so i guess
i have to do atransform somehow to polar coordinates
?===
===
Subject:
: Re: Integration question> How do i integrate
x^(-1/2)exp^(-x) with respect to x and x goes from 0 to>
infinity ? The answer is supposed to be Sqrt[Pi], so i guess i
have to do a> transform somehow to polar coordinates ?Let u =
sqrt(x). Then you'll be integrating exp(-u^2)du from 0 to oo,
which presumably you've seen.===
===
Subject:
: Re: GCD for Rational
Numbers Adjunct Assistant Professor at the University of
Montana.>> So, yes, it is pretty well known, but calling it a
GCD is a rather>> breathtaking abuse of language.>Let
define>GCD(a/b, c/d) = GCD(a,c)/LCM(b,d)>LCM(a/b, c/d) =
LCM(a,c)/GCD(b,d)>It's easy to verify that they are
distributive>GCD(m/n * a/b , m/n * c/d) = m/n * GCD(a/b,
c/d)This is a misnomer again. An operation is not
distributive. Theproper use of the adjective is to say that
one operation distributesover another. In this case, what you
are saying is that if weconsider your GCD as a binary
operation on rationals, thenmultiplication distributes over
GCD (and over 'LCIM')>associate, commutative and satisfy
absorption law>LCM(a/b, GCD(a/b, c/d)) = a/b>(These are
properties from Mathworld page). I seem to have trouble
with>Knuth identity, however,>GCD(2^a/b-1, 2^c/d-1) =
2^GCD(a/b,c/d) -1GCD is not defined for these animals, unless
b|a and d|c, because you onlydefined it for rational numbers.
2^{a/b} is irrational unless b|a.--
==============================================================
========It's not denial. I'm just very selective about what I
accept as reality. --- Calvin (Calvin and
Hobbes)=======================================================
==================
===
Subject:
: Re: GCD for Rational Numbers
Adjunct Assistant Professor at the University of Montana.>> In
fact, it is simply the largest common measure of teh two
rationals;>> writing a = p/q and b=r/s with
gcd(p,q)=gcd(r,s)=1, then any common>> measure will be of the
form a/k=b/m for some integers k and m. Thus,>> its reduced
numerator will be a divisor of both p and r, hence of the>>
gcd of p and r; and the reduced denominator will be a multiple
of both>> q and s, hence of their lcm. So any common measure of
the form u/v,>> with gcd(u,v)=1 will have |u|<= gcd(p,r) and
|v|>=lcm(q,s), and so>> will be smaller than the common
measure found.>.....>> If you change to lcms, then from p/q
and r/s you get the rational that>> has gcd(q,s) as
denominator and lcm(p,r) as numerator. This would be>> the
smallest rational that has both a and b as measures (the
smallest>> rational measured by both a and b).>> For suppose
that the rational u/v is measured by both a=p/q and>> b=r/s.
Then ka = mb = u/v for some integers k and m. The reduced
form>> of ka will have a denominator that divides q and
numerator which is a>> multiple of p; and the reduced form of
mb will have a denominator that>> divides s and a numerator
that is a multiple of r. Thence u/v will>> have u a multiple
of lcm(p,r) and a dneominator a divisor of>> gcd(q,s). So any
rational measured by both a and b will be of the form>> u/v
with lcm(p,r)<=u and gcd(q,s)>=v, hence lcm(p,r)/gcd(q,s) will
be>> the smallest such rational.>Let's call proposed extension
for rational numbers as GCDR (LCM,>correspondingly).>Are you
suggesting that>GCDR(a/b, c/d) = GCD(a,c)/LCM(b,d)>GLCM(a/b,
c/d) = LCM(a,c)/GCD(b,d)I'm not suggesting. That ->IS<- your
definition. Provided gcd(a,b)=gcd(c,d)=1.>? This clearly
doesn't fit my definition.
Consider>GCDR(87/22,9/5)=>=GCDR(2^-1*3^1*11^-1*29^1,3^2*5^-1)=
>=2^min(-1,0)*3^min(1,2)*5^min(-1,0)*11^min(-1,0)*29^min(0,1)=
>=2^-1*3^1*5^-1*11^-1=3/110>whereas>GCD(87,9) = 3*29Really?
You think that 29 divides 9? How on Earth do you manage
that?>LCM(22,5) = 110--
==============================================================
========It's not denial. I'm just very selective about what I
accept as reality. --- Calvin (Calvin and
Hobbes)=======================================================
==================
===
Subject:
: Re: GCD for Rational
Numbers>GCD(87,9) = 3*29> Really? You think that 29 divides 9?
How on Earth do you manage that?I apologyse for trigger happy
posting (which provokes trigger happyreplies?) I hope I didn't
discourage you from following other messages inthe
thread.===
===
Subject:
: Re: expectation of the product of two
dependent random variables>Hi sorry to bother you guys
here.>Let random variable X = a_1r_1+a_2r_2+...+a_nr_n, and
>random variable Y = b_1r_1+b_2r_2+...+b_nr_n, where a_i and
b_i>(i=1...n) are>constants, and r_i is i.i.d. random variable
chosen from N(0,1). Note>that r_i in both X and Y are the
same.>So how to compute the expectation of exp(p times X times
Y), that>is>E[exp(p times X times Y)] = ?>> X and Y are jointly
normal random variables with means 0, variances>> E[X^2] =
sum_i (a_i)^2 and E[Y^2] = sum_i (b_i)^2, and covariance>>
E[XY] = sum_i a_i b_i. By diagonalizing the covariance
matrix>> ( E[X^2] E[XY] )>> ( E[XY] E[Y^2] )>> you can reduce
to the case where n=2. >Thank you very much, Robert.>I am
sorry but I didn't get your point. Given the covariance
matrix, >how do you compute the E[e^(pXY)] where p is a
constant?If Z is a normal vector with mean O and covariance
matrixSigma, and Z'AZ, A symmetric, is a quadratic form,
E(exp(Z'AZ)) = det(I - 2*A*Sigma)^(-.5),assuming that all
characteristic roots of A*Sigma have real parts less than .5;
otherwise the expectation doesnot exist.-- This address is for
information only. I do not claim that these viewsare those of
the Statistics Department or of Purdue University.Herman
Rubin, Department of Statistics, Purdue
Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX:
(765)494-0558===
===
Subject:
: Re: the imaginary part of Fourier
transform of u(t) exp(-at^2)Gary and Bob,Haifang> I am trying
to understand the fourier transform of gaussian
function.Sorry> if this is too simple for you.> A fourier
transform of a gaussian function exp(-at^2) is a
gaussianfunction> with imaginary part always zero. What if for
a function as {exp(-at^2), if> t>=0 and 0 if t<0). (This is
usefull in NMR measurement. We can onlymeasure> from time 0
and up.)> I have tried with excel and other FFT software. The
real part is still> gaussian but the imaginary is not zero any
more. My question is whether we> can express the imaginary part
in any math expression?> Many thanks===
===
Subject:
: Re: Euclid
Algorithm for GCD(a/b,c/d) Adjunct Assistant Professor at the
University of Montana.>GCD(87/22,9/5) = ?You can simply follow
Euclid's original, since his algorithm works forany two
commeasurable quantities, i.e., quantities having a
commonmeasure (which is the case with the two rationals).First
find a common measure, so that both numbers can be expressed
asinteger multiples of that measure. The easiest way to do so
is to take1/b*d as a common measure. Thena/b = a*d[1/bd]c/d =
c*b[1/bd]and you havegcd(a/b, c/d) = gcd(ad(1/bd),cb(1/bd)) =
(1/bd)*gcd(ad,cb)and the latter is a gcd of integers, which
can be solved using theusual method.Likewise, to find your
lcm, we find the lcm of ad and cb (which isequal to
adcb/gcd(ad,cb)), and then divide by bd.--
==============================================================
========It's not denial. I'm just very selective about what I
accept as reality. --- Calvin (Calvin and
Hobbes)=======================================================
==================
===
Subject:
: Re: Euclid Algorithm for
GCD(a/b,c/d)>GCD(87/22,9/5) = ?> You can simply follow
Euclid's original, since his algorithm works for> any two
commeasurable quantities, i.e., quantities having a common>
measure (which is the case with the two rationals).That is
what I thought I did in the root message. (The folloup
messageshould have <snipped> mark to avoid confusion.)> First
find a common measure, so that both numbers can be expressed
as> integer multiples of that measure. The easiest way to do
so is to take> 1/b*d as a common measure. Then> a/b =
a*d[1/bd]> c/d = c*b[1/bd]> and you have> gcd(a/b, c/d) =
gcd(ad(1/bd),cb(1/bd)) = (1/bd)*gcd(ad,cb)> and the latter is
a gcd of integers, which can be solved using the> usual
method.Distributive property allows reducing the problem from
rationals tointegers (not to mention alternative
straightforward method that leveragesformula gcd(a/b, c/d) =
gcd(a,c)/lcm(b,d)). The message in the root of thethread
simply demonstrated that Euclid algorithm works directly
withrationals (which might simply be a different wording of
your common measurecomments).===
===
Subject:
: Re: Euclid Algorithm
for GCD(a/b,c/d) Adjunct Assistant Professor at the University
of Montana.Distributive property allows reducing the problem
from rationals to>integers (not to mention alternative
straightforward method that leverages>formula gcd(a/b, c/d) =
gcd(a,c)/lcm(b,d)). The message in the root of the>thread
simply demonstrated that Euclid algorithm works directly
with>rationals (which might simply be a different wording of
your common measure>comments).Sorry; the original message
showed up corrupted in my screen, withthings like a/b - 2c/d
or whatever.But, yes: Euclid's algorithm shows up in the
Elements applied toarbitrary co-measurable quantities x and y.
You find the largestinteger N such that Nx is no larger than y,
and then show that thelargest common measure of x and y is the
same as largest commonmeasure of x and y-Nx. Lather, rinse,
repeat.--
==============================================================
========It's not denial. I'm just very selective about what I
accept as reality. --- Calvin (Calvin and
Hobbes)=======================================================
==================
===
Subject:
: Re: Euclid Algorithm for
GCD(a/b,c/d)> GCD(87/22,9/5) = ?Multiply by a common
denominator (in this case 110, need not be LEASTcommon
denominator): 87/22*110 = 435, 9/5*110 = 198Compute gcd using
integer method: GCD(435,198) = 3Divide by that same common
denominator: 3/110, answer.-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re: Euclid
Algorithm for GCD(a/b,c/d)> GCD(87/22-2*9/5) =
?-----^^^^^^^^^^^^^^GCD(87/22, 9/5) = ?I really need
copy-and-paste buggyproof checker.===
===
Subject:
: Re: Euclid
Algorithm for GCD(a/b,c/d)>GCD(87/22-2*9/5) = ?>
-----^^^^^^^^^^^^^^> GCD(87/22, 9/5) = ?> I really need
copy-and-paste buggyproof checker.I think you got
it.3/110===
===
Subject:
: Exercise on proper classes in ZFI have to
solve following exercise, i.e. to prove that following classes
areproper classes:1 - The classe of all powersets2 - The classe
of sets containing a given set a3 - The classe of equivalence
relationsPlease help giving some hints.Regarding 1, I'm able
to derive the fact (just using some definition...)that if the
given classe is a set A, then the powerset of A belongs to
A,which seems strange, but I wasn't able to derive a
contradiction.Regarding 2, I'm able to derive that if the
classe is a set b then: a in b<--> a in a <--> b in b. Then,
so what ?===
===
Subject:
: Re: Exercise on proper classes in ZF
Adjunct Assistant Professor at the University of Montana.>I
have to solve following exercise, i.e. to prove that following
classes are>proper classes:>1 - The classe of all powersets>2 -
The classe of sets containing a given set a>3 - The classe of
equivalence relations>Please help giving some hints.>Regarding
1, I'm able to derive the fact (just using some
definition...)>that if the given classe is a set A, then the
powerset of A belongs to A,>which seems strange, but I wasn't
able to derive a contradiction.If the power set of A is an
element of A, since A is an element of thepower set, this
would contradict Foundation, would it not?--
==============================================================
========It's not denial. I'm just very selective about what I
accept as reality. --- Calvin (Calvin and
Hobbes)=======================================================
==================
===
Subject:
: Re: Exercise on proper classes in
ZF>>I have to solve following exercise, i.e. to prove that
following classes are>>proper classes:>>1 - The classe of all
powersets>>2 - The classe of sets containing a given set a>>3
- The classe of equivalence relations>>Please help giving some
hints.>>Regarding 1, I'm able to derive the fact (just using
some definition...)>>that if the given classe is a set A, then
the powerset of A belongs to A,>>which seems strange, but I
wasn't able to derive a contradiction.>>If the power set of
A is an element of A, since A is an element of the>power set,
this would contradict Foundation, would it not?Yes, and
Foundation is in ZF. Personally, though, I prefer proofs which
avoid it if they can, since Foundation is irrelevant to almost
all of mathematics. (Got that from Kunen.) How about showing
that if there exists a set of all power sets, then there
exists a set of all sets?-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu===
===
Subject:
: Re: Exercise on proper
classes in ZFZF has no proper classes. What system are you
REALLY using?===
===
Subject:
: Re: Exercise on proper classes in
ZF>ZF has no proper classes. What system are you REALLY
using?> We've been through this recently. Formally, one can
consider a class to be a formula. See, e.g., Kunen. E.g., The
ordinals form a proper class translates as There does not
exist x such that for all y y is in x if and only if y is an
ordinal.-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu===
===
Subject:
: Re: P, NP and
coNP>[...] Tell>us how you can establish (coNP subsetof NP)
==> (coNP = NP)[...]>Let L be any language in NP. The
complement(L) is in coNP by>definition. If (coNP subsetof NP),
then it follows that>complement(L) is in NP. But then
complement(complement(L))=L is in>coNP.> What you said is
correct and anyone with half a live brain cell should> be able
to see that point. But you still did not do it so
well.Newstome's proof is fine.> [...] But give me something
better, please!What newstome gave is what tcne requested.>
With this established, the only triviality left is to show
coNP> subsetof NP. If you really want to do math, we can
finish the job (of> explaining something that needs hardly any
explanation).> There are two simple ways of doing it. One is to
do a direct> simulation of M2 for M. The other is for you to
point out exactly the> spot in the definition/specification of
M2 that fails to simulate M,> or fail to halt (within the
specified polynomial time) in a rejecting> configuration (when
x notin L is the input, where L is the NP> language in
question). Tell us, which way you want it and I will>
oblige.The un-fixable error in the 'proof' is the false
assertion near thetop of page 4: If we switch labeling of the
rejecting and accepting states, M^2 becomes a
non-deterministic Turing machine that accepts L' and rejects
L, both in polynomial time.For a counter-example, see Rune
Bang Lyngsoe's reply to rosiin these same groups on
1998/12/09, message ID:<74mt9p$fnr$1@xinwen.daimi.au.dk>#1/1,
currently available at:
http://www.google.com/groups?selm=74mt9p%24fnr%241%
40xinwen.daimi.au.dktcne is obviously the same author as rosi.
His current demonstration that NP=coNP is essentially the same
as the argument he posted under NP=coNP Made Easier? over five
years ago. A number of replies showed it wrong and explained
why, and the current version does nothing to fix the key
error.-- --Bryan===
===
Subject:
: Re: New Knowledge>How, when and
why does *New Knowledge* come to humankind? I screwed this
thread up awhile back by posting a response only tosp, thereby
creating a whole new thread. Plus my answer quicklydeteriorated
into chest banging I believes and obvious facts. Thishas more
to do with my crappy (but hopefully improving) usenet
writingstyle than what I was trying to say. Please allow me to
simplify: It's safe to say we are composed of the stuff of the
universe, like atree is composed of the stuff of the seed from
which it grew. Newknowledge, or adaptability in the case of the
tree, is encoded in theseed. Taking the big bang as our seed,
new knowledge (all possibleknowledge?) would likewise be
encoded in that seed. Over time, allallowable possibilities up
to a given moment might add up to aproduct we call discovery.
The idea is inherent in the word;dis-cover, not create. So I'm
thinking maybe, How - It is encoded in the stuff of which we
are made When - At the moment when all allowable possibilities
add up to aparticular discovery Why - Since we are of the
universe, and we ourselves have a deepinterest in
self-knowledge, the purpose of the universe might
beself-knowledge. The above might be old hat, and probably
*is*, however I'm not thatwell read, so I had to come up with
it myself. Cookie, please. I hope I have made myself clear,
for what it's worth. I find itdifficult to translate stuff in
my head in such a way as to make itclear to someone who is not
in there with me. Have I made that clear? ===
===
Subject:
:
Commutative algebraHi all,does anyone know a good and
entertaining math book on commutativealgebra/ring theory ? I
went through Neal McCoy 'Rings and ideals' andfelt in love
with it. It is both stimulating and progressive in theapproach
and i strangely (i actually don't know why) find
itentertaining/amusing.But it lacks exercices (potentially
with solutions/hints) tostrengthen up what had might have been
loosely understood throughoutthe text and a more advanced look.
My aim would be to be as wellprepared as possible to dig into
books like Miles Reid's'Undergraduate commutative algebra'
(see thread called Pleaserecommend some entertainment math
books) or Matsurama's book oncommutative algebra. The few
first paragraphs of Reid's book are quiteappealing to me
(local algebra, ...).In an other subject, do you have any
comments on Humphreys's'Introduction to lie algebra' ? Is it
accessible to an undergraduateself-study folk ?Thank you for
your advices,Charles===
===
Subject:
: ENAR job placementThere will
be the ENAR meeting in Pittsburgh, PA (March 28 - 31,attend
the previous ENAR job placement before? Can you provide
anyexperience such as how many positions can I
expect?Zhu===
===
Subject:
: My brother's puzzleHiMy brother has
shown me this puzzle
-http://www.rabid.oneuk.com/ultimategame.htmland I have no
idea what to do? He said it is based on a
mathematicalprinciple, and I'm pretty hot at math but this is
just frustrating!Help?Greta ;)xxx===
===
Subject:
: Re: My brother's
puzzle@posting.google.com:> Hi> My brother has shown me this
puzzle -> http://www.rabid.oneuk.com/ultimategame.html> and I
have no idea what to do? He said it is based on a
mathematical> principle, and I'm pretty hot at math but this
is just frustrating!> Help?> Greta ;)> xxxJust looking over
the source code, unless I missed something, the only possible
response is Sorry, it's not THERE. Game Over. Then there is
what appears to be a lot of garbage thrown around in in the
source also. In fact, taking out the comments, it looks like a
way to increase a hit counter of some sort. -- ===Comments:
This message probably did not originate from the above
address. It was automatically remailed by one or more
anonymous mail services. You should NEVER trust ANY address on
Usenet ANYWAYS: use PGP !!! Get information about complaints
from the URL belowX-Remailer-Contact: http://80.65.224.85/POL/
In case my abuse address is unreachable: It is because it has
been flooded by <mgg@uiuc.edu>, please contact
<abuse@uiuc.eduX-Mail2News-Contact: http://80.65.224.85/I am
from Niger. The anglo-saxon race steals our yellow cake at
dirt cheapprices. We are not allowed to buy uranium enrichment
equipment tomake our uranium expensive and get the compensation
we deserve. The wholeworld runs on our uranium, the african
uranium of the black people ofNiger.I am sick of this
**multi-dimensional injustice**.That is why I tell the world,
my black, african christian, muslim,indian, south african,
filipino, east asian, thai brothers. *Put morality aside
forawhile and follow strategy*.Put morality aside forawhile
and follow strategy -- for the ultimatemoral victory.Do not
cede yankee an iota of support on gay-lesbian issue.
Bargainwith him the right to polygamy that is part of your
culture andadjustment in need. Look, you do not cheat. You do
not go aroundscrewing women out of wedlock. You are also
immigrant. At some stageyour children will marry out of race
and will also want to marryin-race. ** You need polygamy for
your identity and culture **. It isyour birth right. It may be
your survival. Possibly your wifeis not fertile and you love
her. You do not want to adopt children.I tell you this fact. I
must tell you because it is * a basis for the case* forour
rights. I must make it public.My daughter was an undergrad
student at University of Florida in Gainesville. She was in
the materials science department. There is aprofessorthere by
the name of Eugene Goldberg. He is a rich jew. He has two
wives.Onein Alachua county and one in Atlanta, Georgia. The
whole of the UNIVERSITY OF FLORIDA materials science
department knows eugene goldberg has twowives. Dr. Eugene
Goldberg's first wife lives in Florida, the secondin Georgia.
He is rich enough to fly back and forth every week orother.
Department chairman, Reza Abbaschian gave him this position to
get him in the department. Eugene Goldberg is reputed to have
100+ patents onplastics when he worked for GE. I have not
certified these facts on patents. But my daughter told me that
she spoke to both the wives for she had to talk to the prof
when he was at one home or another and called boththe numbers,
and spoke to both the wives. Dr EG has income in top 100people
of the state of florida, was a rumor I once heard. Possibly
true.So why can they make exception for the jew and the mormon
to polygamy?and not for the black, african, the muslim, the
hindu, the sikh andthe Chinese? The white can cheat on his
wife. Clinton did it manytimes. Why must the white anglo
christian impose his view on us? Weare free men and we have
our own code of marriage.So why can they make exception for
the jew and the mormon to polygamy?and not for the black,
african, the muslim, the hindu, the sikh andthe Chinese? The
white can cheat on his wife. Clinton did it manytimes. Why
must the white anglo christian impose his view on us? Weare
free men and we have our own code of marriage.Bargain to make
it illegal to ask at INS interview, to any hindu,muslim, sikh,
african or any whose native culture has polygamy. Setaside your
own religous views or whatever and support the gay till
themajority of white give you the right to your way of life
andtradition.One thing we learnt at Cancun is that the white
man bargains veryhard. We must do the same. White man believes
that the whole universeis centered around him, and sun revolves
around him. That is theirchristianity. We must return to our
animist traditions with science,like the chinese, japanese and
the hindu. Christianity belong to thewhite man, but not
science.The essential issue is this: We must drive the **
immigration laws ** to make it illegal and unconstitutional to
ask opinion of the hindu,muslim, sikh, african or any whose
native culture has polygamyregarding his belief in that or any
conversation about that.Remember, the white and the jew think
us inferior. One has superioritycomplex from the skin and the
other has from their talmud.Remember, the white and the jew
think us inferior. One has superioritycomplex from the skin
and the other has from their talmud.Their elite believes us
inferior. Shockley said: Blacks are aninferior race. He
offered money to get castrated the blacks or undergovasectomy.
He had a flowery language for it which is what yankee hasfor
all immoral and criminal activity to another race, nation
orreligion.We must not be happy with the words of massage. We
must get it writtenin the law. We must get the immigration law
to open door for polygamyand racial bce. George Bush and the
Repuglican party as well asRepuglicans disguised as
demoncrats, essentially chameleons - are *all*racists
inside.In the matter of definition, we must get this in the
law. Absolutely nodiscrimination in any way on polygamy for
those who have had historicalpolygamy. I have been reading a
lot of books on polygamy. I know it is a sikh tradition. Their
kings have many wives.I want hindus, sikhs, muslims, arabs,
africans all to be able to practicepolygamy freely per their
needs like the jew and the mormon. The whiteas we know likes
to scrounge around and does not need polygamy, andrarely
ashamed of cheating or infidelity. Clinton did it. Bush
hadmoney to divorce his first wife and keep her silent.I want
hindus, sikhs, muslims, arabs, africans all to be able to
practicepolygamy freely per their needs like the jew and the
mormon.We must recall, we did not win our rights by hiring
lawyers and in thecourts. We won by the civil war. It was not
given us by asking anddelivery.It was gained by active
resistance.Forget about 911 . This is not just a distraction
from the realproblems, it is a B-I-G B-I-G D-E-C-E-P-T-I-O-N
FROM THE REALPROBLEMS by white. It is a dark and evil smoke
screen. You are the victim.Those who died cannot be brought
back. Yank ruling elite provokethe world to no end to invite
this retaliation from desperate menwilling to die in suicide.
Explain who commits suicide? The desperateor mentally ill.
Well, racism and victimization makes its victimsmentally ill
as well.911 is a dark and evil smoke screen. It is
inconsequential. It does notmatter in our lives today. What
matters is that we fix our foreign policiesand stop the flow
of our aid to israel and ask for payment of loans to us.We
need passion and anger to win our legitimate rights. We need
astrategy.I say this to you: fight for your rights, and
support the gay, opposethe white sensibility till they accept
our sensibility on polygamy.It is OK to say: I will support
the gay for purely strategic and politicalreasons. To avoid
argument, I suggest, we start a signing list. The numberswill
tell the whole game. Either white give us the right to
polygamy in USAor we deny them the right to their definition
of marriage. We must not letthem use us. We must not let them
use us. We must not let them use us. We must not letthem use
us.Remember the name, Eugene Goldberg has two wives, on in
alachua, one inatlanta.We need that right as well. I want have
wife in NY and want one from Niger.Indians wantone wife in US
and one in India. Arabs want one wife in US and one inArabica.
AlsoAfricans like me have stronger drives.I believed in
monogamy. Now for me it has become a political issue
ofrights.I have evolved.Support us even if you are not a
polygamist. It is an issue of rights, astick to beat theyank
with -- and it is very fair.Support us even if you are not a
polygamist. It is an issue of rights, astick to beat theyank
with -- and it is very fair.===
===
Subject:
: What does this
mean?(logic in set theory)I've purchased a book recently-N.
Bourbaki's Set Theory.When I read through the first chapter,
I've turned down.At first, I didn't understand what this means
for logic signs-tx(A)(tis not t-it's tau, you know.).The book
says it has the following meaning. (tx(R)x)R <-- there exists
x such that R.( is vertical bar.)Is there anyone who knows
what it means?It didn't appear in the other textbooks for set
theory.They just started with quantifier signs.===
===
Subject:
: Re:
What does this mean?(logic in set theory)> I've purchased a
book recently-N. Bourbaki's Set Theory.> When I read through
the first chapter, I've turned down.Well, it is not exactly
known as the best presentation of Set theory (oranything else)
for beginners :-)> At first, I didn't understand what this
means for logic signs-tx(A)(t> is not t-it's tau, you know.).>
The book says it has the following meaning.> (tx(R)x)R <--
there exists x such that R.( is vertical bar.)> Is there
anyone who knows what it means?This is known as Hilbert's tau.
It is a universal choice function: ifthere is an x such that
P(x), then tau_x P(x) is such an x.> It didn't appear in the
other textbooks for set theory.> They just started with
quantifier signs.Yes. The main reason is that now the Axiom of
Choice is an (easy) theoremfrom the rest of ZF and logic
(predicate) calculus. This idea seems to havebeen abandoned
(even if G.9adel's works shows that it is not more
powerfulthan AC)===
===
Subject:
: Finding roots of an entire
function charset=Windows-1252Consider the following Maple
code:N:=6;for i from 1 to N dof[i]:=z;od:F:=proc(n)if n=1
thenexp(f[1]);elseexp(f[n])^F(n-1);fi:end:HLW:=proc(y,m,n)
local
s,p,sol,i,aprx,dy,dist,solution;dist:=infinity;s:=series(z*F(n
),z,m);p:=convert(s,polynom);sol:={fsolve(p=y,z,complex)};for
i from 1 to nops(sol)do aprx:=evalf(subs(z=op(i,sol),z*F(n)));
dy:=evalf(abs(y-aprx)); if dy<dist then solution:=op(i,sol);
dist:=dy; fi;od;solution;end:The code above finds an
approximation to a, such that: z*F(n,a)=y,
whereF(n,z)=(e^z)^(e^z)^...^(e^z) {n-terms},by going through
all the roots of the equation T_m(z)=y, where T_m(z) is
theTaylor Polynomial of z*F(n,z) of order m.I am trying to
understand why it works. Have I got it right?Proof:Let
g(z)=z*F(n,z)-y, andg_k(z)=T_k(z)-y, with T_k(z) being the
Taylor Polynomial of order k ofz*F(n,z).g_k(z)->g(z)
uniformly, and all functions are entire.In particular g(z) is
entire and non-constant, so according to Picard'sLittle
Thereom,=(Ay)(with at most one exception)(Ea):
a*F(n,a)=y,=>(Ea): g(a)=0.Once we have a root a, Hurwitz's
Theorem:http://mathworld.wolfram.com/
HurwitzsRootTheorem.htmlprovides for a neighborhood
|x-a|<delta, and a number m=N,such that within the
neighborhood |x-a|<delta, =(Ak>m=N):g_k(z)=0 has a root in
that neighborhood.In particular: (Ea_T: T_n(a_T)=y, and a_T
satisfies:|a_T-a|<delta.Using the continuity of
z*F(n,z),=|y-a_T*F(n,a_T)|<epsilon, and the code above picks
an a_T that minimizesepsilon.Hurwitz's Theorem doesn't say
anything about how large m=N should be or whatdelta(or epsilon
is) is, but it appears as though if I increase m, I getbetter
and better approximations for a. Why is this so?Is there a
better choice for the code to pick between z and
conjugate(z),which both minimize epsilon, so as to make the
code more well-defined?Also, according to Picard's Little
Theorem, it appears as though the codeabove will work for all
y, with at most one exception, but one should bereally unlucky
to hit this y. (if it exists. I know it doesn't exist
forz*exp(z) for example).--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
understandable===
===
Subject:
: Re: Finding roots of an entire
function charset=utf-8 Ioannis <morpheus@olympus.mons>
[CapitalEth][EDoubleDot][Micro] .b3 .b9[EDoubleDot].b9> The
code above finds an approximation to a, such that: z*F(n,a)=y,
whereSorry, typo. Should read:> The code above finds an
approximation to a, such that: a*F(n,a)=y, where--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
understandable===
===
Subject:
: Re: Proving associativity: But, what
do you do when all that you have is a table? Is there: anything
other than exhaustive brute force available? The book: that I'm
using as a text (Landin) poses this as a question; one: which
I've never been able to answer. Commutativity is easy to: see
from a table -- just look for symmetry about the major
diagonal.: Is there a simple way to check for
associativity?Not really. That's why groups are not usually
defined by tables, but assome kind of symmetries of an object,
in which case the associative law isobvious.Quoting from a
sci.math post by Dave Rusin
at:http://www.math.niu.edu/~rusin/known-math/97/numgrps4 I
hate to discourage anyone working on a mathematical project,
but I have to say that enumerating groups by creating lots of
Cayley tables ... is really kind of pointless. This tends to
treat the associative law as sort of a nice accident when it
happens, rather than a key feature of group theory.Ted
Hwa===
===
Subject:
: Re: Proving associativity>Back in 2002, I
posted a question about how to prove that all groups>of size
<= 5 were Abelian. With the hints that I received then
(right>before losing my Usenet access), I made some progress.
I had thought>that I had proved the cases for groups of size
1, 2, and 3. However,>I just realized that there was a biiig
hole in my proofs. I hadn't>shown that the objects with which
I was working had the associative>property, and therefore had
no right to claim that they had anything>to do with groups!> I
don't know how much detail you're looking to include in your
proof,> but this problem really isn't too hard.> For instance,
every group of prime order is cyclic, hence Abelian. So> the 2,
3, 5 cases are taken care of. The 1 case is obvious. So now>
all you have to take care of is the order 4 case. To do this
quickly,> you'll probably want to look into the Sylow
p-subgroup theorems.> If you want to prove this using more
elementary facts, you've got your> work cut out for you.> 'cid
'oohTheorem 1 (Lagrange's Theorem). The order of a subgroup of
a givenfinite group divides the order of the group. In
particular, the orderof an element divides the order of the
group.Theorem 2. A subgroup of index 2 is normal.Definition.
If H and K are subsets of a group G, then HK = {hk : h in H,k
in K}.Theorem 3. If H and K are subgroups of a group G, then
|HK| = (|H| |K|)/(|H intersect K|).Theorem 4. If H and K are
normal subgroups of G, and the intersection of Hand K is {1},
and HK = G, then G is isomorphic to H x K.Theorem 5. All
cyclic groups are abelian.Theorem 6. All groups of order 4 are
abelian.Proof:Let G be a group with order 4, and denote its
identity by 1. Let x bea nonidentity element of G. Then by
Theorem 1, x has order 2 or 4.If x has order 4 then G is
cyclic and hence abelian by Theorem 5. Suppose x has order 2.
Let y be an element of G not equal to 1 or x. Then y has order
2 or 4.If y has order 4, then G is cyclic and hence abelian.
Suppose y has order 2.Then {1,x} and {1,y} are normal
subgroups of G by Theorem 2,and their intersection is {1}.
Also, |HK| = 4 by Theorem 3. So, HK = G.Therefore, G is
isomorphic to {1,x} x {1,y} by Theorem 4. But {1,x} and{1,y}
are abelian groups and the direct product of abelian groups
isabelian (commutativity is inherited due to the componentwise
operation onthe direct product). Therefore G is abelian.
Q.E.D.No Sylow theorems (they wouldn't say much for a group of
order 4 anyway).OP:The point is that showing that all groups of
order less than 6 are abelianis an easy exercise once you've
built up enough machinery, and I don't thinkit's very
instructive to do this exercise after only just learning what
a group is (though you could probably write a computer program
to solve it for you). Leonard Blackburn===
===
Subject:
: Re: Proving
associativity>> Back in 2002, I posted a question about how to
prove that all groups>> of size <= 5 were Abelian. With the
hints that I received then (right>> before losing my Usenet
access), I made some progress. I had thought>> that I had
proved the cases for groups of size 1, 2, and 3. However,>> I
just realized that there was a biiig hole in my proofs. I
hadn't>> shown that the objects with which I was working had
the associative>> property, and therefore had no right to
claim that they had anything>> to do with groups!I don't know
how much detail you're looking to include in your proof,>but
this problem really isn't too hard.>For instance, every group
of prime order is cyclic, hence Abelian. So>the 2, 3, 5 cases
are taken care of. The 1 case is obvious. So now>all you have
to take care of is the order 4 case. To do this
quickly,>you'll probably want to look into the Sylow
p-subgroup theorems.So what is the connection between the
Sylow p-subgroup theorem andproving that groups of order 4 are
abelian?Derek Holt.===
===
Subject:
: Re: Proving associativity> Back
in 2002, I posted a question about how to prove that all
groups> of size <= 5 were Abelian. With the hints that I
received then (right> before losing my Usenet access), I made
some progress. I had thought> that I had proved the cases for
groups of size 1, 2, and 3. However,> I just realized that
there was a biiig hole in my proofs. I hadn't> shown that the
objects with which I was working had the associative>
property, and therefore had no right to claim that they had
anything> to do with groups!>I don't know how much detail
you're looking to include in your proof,>but this problem
really isn't too hard.>For instance, every group of prime
order is cyclic, hence Abelian. So>the 2, 3, 5 cases are taken
care of. The 1 case is obvious. So now>all you have to take
care of is the order 4 case. To do this quickly,>you'll
probably want to look into the Sylow p-subgroup theorems.> So
what is the connection between the Sylow p-subgroup theorem
and> proving that groups of order 4 are abelian?> Derek
Holt.I'm not sure Sylow would help too much here, but try
this:|G| = 4, let Z(G) be the centre of G.Z(G) <= G so |Z(G)|
has size 1,2 or 4 by Lagrange's theorem. If it hassize 4 then
we're done. If it has size 2 then we're done, because|G/Z(G)|
= 2 so is cyclic and therefore G is abelian. So suppose ithad
size 1. Take g != 1 in G. Then if g^2 != 1, g and g^2
commutewhich contradicts |Z(G)| = 1. Otherwise, if we cannot
pick such a g,then g^2 = 1 for all g. If G = {1,a,b,c} then
(ab)^2 = 1 => abab = 1=> ab = ba (since b^-1 = b) and hence G
is abelian, which contradicts|Z(G)| = 1.Therefore any group of
order 4 is abelian.from Jonny!===
===
Subject:
: Re: Proving
associativity Adjunct Assistant Professor at the University of
Montana.>I'm not sure Sylow would help too much here, but try
this:>|G| = 4, let Z(G) be the centre of G.>Z(G) <= G so
|Z(G)| has size 1,2 or 4 by Lagrange's theorem. If it has>size
4 then we're done. If it has size 2 then we're done,
because>|G/Z(G)| = 2 so is cyclic and therefore G is abelian.
So suppose it>had size 1. Take g != 1 in G. Then if g^2 != 1,
g and g^2 commute>which contradicts |Z(G)| = 1.Well, you have
to work a bit more, don't you? So far, g commutes with3
elements, you haven't shown it commutes with all 4 elements.
You arereally using that the ->centralizer<- of g,C_g = {x in
G : gx=xg}is a subgroup, so by Lagrange's theorem it has
either 1, 2, or 4elements, and since it has 3 already, it must
have 4, so C_g = G, andof course, g in Z if and only if C_g =
G.--
==============================================================
========It's not denial. I'm just very selective about what I
accept as reality. --- Calvin (Calvin and
Hobbes)=======================================================
==================
===
Subject:
: conjugate
permutationshttp://math.berkeley.edu/~wodzicki/113.S04/Ex1.
pdfFor problem 1, I got my rho to be, after decomposing into
orbits,1 2 3 4 5 6 7 8 9 10 112 1 9 7 3 6 4 10 11 5 8I can
easily get rho inverse. I even checked my answer by
multiplying rhosigma and rho inverse all out? Yet why did my
grader give me half credit forthis problem? Did I do something
wrong?===
===
Subject:
: Re: conjugate permutations Adjunct Assistant
Professor at the University of
Montana.>http://math.berkeley.edu/~wodzicki/113.S04/Ex1.pdf>
For problem 1, I got my rho to be, after decomposing into
orbits,>1 2 3 4 5 6 7 8 9 10 11>2 1 9 7 3 6 4 10 11 5 8>I can
easily get rho inverse. I even checked my answer by
multiplying rho>sigma and rho inverse all out? Yet why did my
grader give me half credit for>this problem? Did I do
something wrong?Don't know. Depends on what he was expecting
you to do, I guess. Areyou sure you are using the same
convention on multiplication? Thereare two ways to interpret
rho*sigma*rho^{-1}: left to right or rightto left.I get that
sigma = (1,5,11)(2,3,8,9) (4,10,6) sigma' =
(2,3,8)(1,9,10,11)(7,5,6)You give rho =
(1,2)(3,9,11,8,10,5)(4,7) so rho^{-1}=
(1,2)(3,5,10,8,11,9)(4,7)Working right to left,
rho*sigma*rho^{-1} maps 1 -> 2 (what rho^{-1}does) then 2->3
(what sigma does) and finally 3->9 (what rho does). Sothis
permutation will map 1 to 9, etc. So you are using
thatconvention.If we work left to right, on the other hand, we
have 1->2 (using rho),2->3 (using sigma), and 3->5 (using
rho^{-1}), so that 1 maps to 5,which is not what sigma' is
supposed to do.So I would make sure you are both using the
same convention incomposing permutations, and if so, then
(politely!) ask the grader/TA toexplain the marking. It's
certainly possible he/she made a mistake.--
==============================================================
========It's not denial. I'm just very selective about what I
accept as reality. --- Calvin (Calvin and
Hobbes)=======================================================
==================
===
Subject:
: Re: conjugate
permutations>http://math.berkeley.edu/~wodzicki/113.S04/Ex1.
pdf>For problem 1, I got my rho to be, after decomposing into
orbits,>1 2 3 4 5 6 7 8 9 10 11>2 1 9 7 3 6 4 10 11 5 8>I can
easily get rho inverse. I even checked my answer by
multiplying rho>sigma and rho inverse all out? Yet why did my
grader give me half creditfor>this problem? Did I do something
wrong?> Don't know. Depends on what he was expecting you to do,
I guess. Are> you sure you are using the same convention on
multiplication? There> are two ways to interpret
rho*sigma*rho^{-1}: left to right or right> to left.> I get
that sigma = (1,5,11)(2,3,8,9) (4,10,6)> sigma' =
(2,3,8)(1,9,10,11)(7,5,6)> You give rho =
(1,2)(3,9,11,8,10,5)(4,7)> so rho^{-1}=
(1,2)(3,5,10,8,11,9)(4,7)> Working right to left,
rho*sigma*rho^{-1} maps 1 -> 2 (what rho^{-1}> does) then 2->3
(what sigma does) and finally 3->9 (what rho does). So> this
permutation will map 1 to 9, etc. So you are using that>
convention.> If we work left to right, on the other hand, we
have 1->2 (using rho),> 2->3 (using sigma), and 3->5 (using
rho^{-1}), so that 1 maps to 5,> which is not what sigma' is
supposed to do.> So I would make sure you are both using the
same convention in> composing permutations, and if so, then
(politely!) ask the grader/TA to> explain the marking. It's
certainly possible he/she made a mistake.>
==============================================================
========It's not denial. I'm just very selective about> what I
accept as reality.> --- Calvin (Calvin and Hobbes)>
==============================================================
========> Arturo Magidin> magidin@math.berkeley.eduOur class
does right to left? Am I okay with my solution?===
===
Subject:
: Re:
What is the number derivative ???> Yep, may be I made my words
unwanted sharp. I just wondered, that the>
abstract/announcements of the two resources sounded very
similar aboutIt's not easy to find, but the paper is available
on the Web
athttp://web.mit.edu/lwest/wwww/intmain.pdf===
===
Subject:
: Re:
What is the number derivative ???> Among the top ten Intel
Science Talent Search winners just announced is one Linda
Westrick, whose project concerns the number derivative --
which is described in the PR squib as a recently-defined
operation.> Can someone be so kind as to post the correct
definition of number derivative for me?> AsimovNot that this
is what Ms. Westrick is doing, but there is a notion
ofp-derivative (or p-derivation) for a prime p. It is defined
by dN/dp = (N^p - N)/p.dN/dp is an integer, by Fermat's little
theorem. Of course, thisoperation isn't actually a true
derivative, since it doesn't satisfythe product rule. (It's
also not linear, but that's another story!)But it does satisfy
a twisted product rule: d(MN)/dp = ((MN)^p - M^p*N + M^p*N -
MN)/p = M^p * dN/dp + N * dM/dpThat power on the M is what
makes it a twisted product rule.Joe Silverman===
===
Subject:
: Re:
What is the number derivative ???> I can't, but this may be
relevant:>
http://www.math.uwaterloo.ca/JIS/VOL6/Ufnarovski/
ufnarovski.htmlStrangley, there are two other papers in the
same journal on the
samesubjecthttp://www.math.uwaterloo.ca/JIS/VOL6/Cohen/cohen6.
htmlhttp://www.math.uwaterloo.ca/JIS/VOL6/Hare/hare3.htmlwhich
don't reference the Ufnarovski one (or vice versa).(possibly)
deeper things
ishttp://www.mathematik.uni-bielefeld.de/documenta/vol-kato/
kurokawa_ochiai_wakayama.dm.htmlIt gives a vague impression as
being a glimpse of 'mathematics of thefuture'.===
===
Subject:
: Re:
Double-Sum Over Coprime Indexes (puzzle problem)> Evaluate it?
I can't even read the friggin' thing.This is odd, since it has
appeared fine on both MathForum and onGoogle, and it is
unusual for *both* sci.math portals to not mess
upascii-art.Yet somewhere still, apparently, the ascii-art is
messed up anyway.I also write my equations in linear-mode for
those with CRAPPYnewsreaders.(as I did below the ascii-art
version in my original post)Why newsreaders and
internet-portals which are set up to displaysci.math and
rec.puzzles, especially those (like MathForum) which
are*devoted* to diplaying math-oriented posts, ever mess-up
ascii-art isa mystery.As for the puzzle, I will post the
answer in a couple days, if I needto.But I bet someone will
get this relatively easy puzzle before then.Leroy Quet>Hint
below:>I bet this is one of the easiest math puzzle, but it
seems worth posting> anyway.Evaluate: oo oo>--- --- 1>
----------->/ / k j (k+j)>--- --->k=1 j=1>GCD(k,j)=1>In linear
mode:sum{k=1 to oo} sum {j=1 to oo, GCD(k,j)=1}
1/(kj(k+j)).>Leroy Quet>Hint:>|>|>V>|>|>V>|>|>V>|>|>V>the
double-sum evaluates to a rational.>Bigger hint
below:>|>|>V>|>|>V>|>|>V>|>|>V>|>|>V>the double-sum is equal
to an integer.>(if I did my math right...)>Leroy
Quet===
===
Subject:
: Average of product approaches zeta(n)Let
a(p,n) be a nonegative integer wherep^a(p,n) is the highest
power of the prime p which divides n.I get that the limit, for
n = integer >= 2,limit{m-> oo}(1/m)*sum{j=1 to m}
product{p=primes, p|j} (1 +floor(a(p,j)/n))= zeta(n).In
ascii-art mode: m 1 -- ------ | | |a(p,j)| m / | | (1
+|------|) -- p|j |_ n _| j=1approaches zeta(n) as m -> oo.If
so, then does n have to be an integer?What is the limit for n
= any real > 1?Leroy Quet===
===
Subject:
: Re: Going back home
charset=iso-8859-7 Brian Tung <brian@isi.edu>   >Assume one is
dropped somewhere on the vicinity of the Milky way.
Coulda>consistent and practical Mathematical model for
navigating around befound>assuming one has unlimited time to
travel?>
http://astro.isi.edu/games/dimension.html!!!!!Beautiful.>
Brian Tung <brian@isi.edu> The Astronomy Corner at
http://astro.isi.edu/> Unofficial C5+ Home Page at
http://astro.isi.edu/c5plus/> The PleiadAtlas Home Page at
http://astro.isi.edu/pleiadatlas/> My Own Personal FAQ (SAA)
at http://astro.isi.edu/reference/faq.txt--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
understandable===
===
Subject:
: Re: Going back home
charset=iso-8859-7 The Ghost In The Machine
<ewill@sirius.athghost7038suus.net>  >Assume one is dropped
somewhere on the vicinity of the Milky way. Coulda>consistent
and practical Mathematical model for navigating around
befound>assuming one has unlimited time to travel?>Making the
question more specific: Does there exist a
sufficientlyaccurate>(to be practical) Mathematical model that
would allow one to calculateone's>way back to Earth, once one
was dropped, say, near a star which sits6,000>light years away
from Earth?> What information is available to the traveler, and
what equipment?I suppose the traveler can use a very fast
computer (with no limitation onmantissa sizes) and can view
the stars at any angle around the ship andcould also use a
good spectroscope and a good telescope.> Also, there are some
considerations regarding dimensions.> The Milky Way has an
estimated radius of 60,000 light> years, taking the black hole
as the center. This is> 5.67 * 10^20 m. log2(5.67 * 10^20) =
68.94, which means> standard IEEE-754 precision equipment
won't quite cut it,> as it can only handle 52 bits in the
mantissa. Of course> one can shift into another coordinate
system when one gets> close enough to one's destination; the
Earth orbits with a> radius of 1AU = 1.5 * 10^11 meter, which
translates into a> galactic radius of about 3.78 * 10^9 AU,
with log2(3.78 *> 10^9) = 31.8. This is still beyond standard
floating-point> metrics, as one has to compute a square root
of sum of> squares. Pluto is about 36 AU away on average (with
an> eccentric orbit); I'm not sure if 100 AU is close enough,>
or not.That's interesting. It appears that if one needs to
calculate distances on acosmic scale, one has to go beyond
Microsoft. :*)> You might want to clarify your question, but
it is an interesting> problem. :-)This problem came to mind,
from recalling a fantastic rocket simulationapplication for
the Macintosh, by Robert Munafo, called Orion, if memoryserves
right. You were aboard a spaceship and you were allowed to
travelvery close to the speed of light. Robert manages to
calculate constellationshapes from the traveler's perspective
within a radius of 40-45 light years.They do indeed looked
very strange! Most would be non-recognizable, had Robnot put
pointers and names on the view panel.Once you were out there,
you could head back, and do other cool things likeput yourself
in orbit around any existing planet, by approaching any
planetappropriately. I seem to recall that Rob calculated all
the star positionsusing assembly language 3D transforms(!).It
sure sounds kind of funny to me, for us to have this vast
Mathematicalknowledge, which becomes virtually useless when
humans are faced with such aproblem :*(> -- > #191,
ewill3@earthlink.net> It's still legal to go
.sigless.--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
understandableMessage-Id:
<405b7264$0$325$ba620e4c@news.skynet.be===
===
Subject:
: Re: Problem
understanding a proof for compactification>> I'm trying to
understand the proof of lemma 38.1 in Topology from James>>R.
Munkres. They start with a imbedding h from a space X to a
compact>>hausdorf space Z and define then X0 as the space h(X)
and Y0 as the>>closure of X0. Now for the part that I don't
understand, they choose then>>a set A disjoint from X that is
in bijective correspondence k with the set>>Y0-X0, can some
one point me to a proof that such a bijection indeed>>exist?>
Are you asking, how do you know you can find a set which is
both> disjoint from X and in bijective correspondence with
Y0-X0? I.e., are> you asking how can one show that:> Given two
sets A and B, there is always a set C which is (a)> disjoint
from A; and (b) in bijective correspondence with B> ?> First:
given any set A, there is a set c which is not in A.> Proof:
Let c = {x in A: x is not in x}. This is a set; if c were an>
element of A, then either c is an element of c or it is not
an> element of c. If c is not an element of c, then by
construction it> must be an element of c. If c is an element
of c, then by> construction it is not an element of c. This is
a contradiction, so> therefore c is not an element of A. QED>
Now let A and B be sets; we want to show there is a set C
which is> disjoint from A and in bijective corresponce with
B.> (Inspired by a post from Fred Galvin,>
ID:<0311111418320.8057-100000@gandalf.math.ukans.edu>):> Let Z
be the set of all elements of elements of A. Let c be
something> which is not in Z (by the above, c exists). Now
let> C = { {b,c}: b in B}.> First, C is disjoint from A: for
if {b,c} is in C and in A, then c is> an element of an element
of A, hence c would be an element of Z. But> we chose c
specifically so it would not be in Z.> Second, C is in
bijective correspondence to B: define f:B->C by f(b) => {b,c}.
This is clearly surjective; it is also injective. If {b,c} =>
{b',c}, then b'= c or b'=b. If b'=b, we are done; if b'=c,
then {b',c}> has only one element, hence {b,c} has only one
element, so b=c as> well, hence b'=c'=b. In either case, b=b'
so f is injective.> Thus, C satisfies the requirements.> this
was indeed what I was looking for.Marc Mertens===
===
Subject:
: Re:
Happy Pi Day>  Richard Henry <rphenry@home.com>
[NonBreakingSpace].8b.96.87.8c .97.99.95
.93fi.92.9b.93.87>> I rarely post on the news
group, but I read it alot.>> However I just wanted to wish
everyone a Happy Pi Day.> March 14th>> 3/14> gets. The
exactly same message has been posted to sci.astro.amateur,
where> it has generated a lot of responses as well.> Bravo to
whoever thought of it.I'll let you know that I did not post
that message tosci.astro.amateur, if such a post even exists
(which I didn't check).I resent you accusing me of trolling.
You should have compared theauthors of the two posts
first.===
===
Subject:
: Re: Happy Pi Day charset=utf-8 Bobby
<bobbysimpi@yahoo.com> [CapitalEth][EDoubleDot][Micro] .b3
.b9[EDoubleDot].b9sci.math>gets. The exactly same message has
been posted to sci.astro.amateur,where>it has generated a lot
of responses as well.>Bravo to whoever thought of it.> I'll
let you know that I did not post that message to>
sci.astro.amateur, if such a post even exists (which I didn't
check).> I resent you accusing me of trolling. You should have
compared the> authors of the two posts first.Mea Culpa. The
authors have different IP's.My apologies :*)--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
understandable===
===
Subject:
: Re: So WTF, I think you're the
Truman too> A-hem. What's SCIENCE got to doo with it?I think
chemical imbces might have a LOT to do with it.===
===
Subject:
: Re:
So WTF, I think you're the Truman too> OK, maybe
alt.sci.physics can answer, if they're not all paralised by
tadchemism.> Feasability analysis shouldn't be difficult.>
One, it *covers* the city, it also moves to different cities
when I'm there, its range> might only be one hectare. Its
always on in within my perimeter.And apparantly no one but you
can hear it.> weather satellites. An array of pulsed laser
90GHZ radar *surveillance* satellites> that see through
clouds.Clouds are the clue. Not buildings, not metal, not
tracking a sub-metre moving watery target, not detecting
oscillations in materials caused by sound vibrations.> Could a
satellite point a laser down and generate a speaker effect in
the air around us?Nope.> At 1st I thought infra red heating
the air at 20hz to 20khz modulation. But it goes through>
buildings aswell as in the open.Nope it doesn't. If that's
your worry, wear a tinfoil hat.> Three, we see through the
atmosphere with our naked ours, its not that opache.???> Four,
class 4 is defined as the minimum power to ignite chemicals,
its only about 1 Watt,> 500mW if I remember, night clubs use
10Watt, of course a satellite will use class 4, I> own a class
3B.???> Five, radio frequency laser beam will go straight
through the eyes whatever class and is no risk.A radio
frequency laser beam? Interesting. Perhaps a definition of
laser is in order:http://en.wikipedia.org/wiki/Laser> Six, I
don't think its ionisation but good for a brainstorm.???>
Seven, modulating a laser at audio frequencies is trivialNope.
Sound is a different kettle of fish altogether.> At the end of
it, assuming the laser is efficient, the power requirment is
bound by the> power requirments of the equivalent speaker that
it emulates. Maybe 1000W. Within> the realms of a satellites
solar panel, or it could have a big pluto stash on board.>
HercIn summary you're desperately trying to rationalize the
ludicrous and impossible.Why is it you leap for the ridiculous
explanations when one a simple one is so obvious to the rest of
us. Just go and see a psychiatrist why don't you?Stop whining
to the disinterested and just do it. If you're so sure of your
sanity, what have you got to lose?===
===
Subject:
: please show
meCould someone please show me this statement:Question: If p
is a prime, then show that o(p^2)=p^2-p.the o in front of the
equation is supposed to denote the Euler function.===
===
Subject:
:
Re: please show me> Could someone please show me this
statement:> Question: If p is a prime, then show that
o(p^2)=p^2-p.> the o in front of the equation is supposed to
denote the Euler function.I'll avoid the obvious joke
response, and assume you know hardlyanything about the
function phi, Euler's totient function. Given a natural number
n, the value phi(n) is defined as the number of natural numbers
less than n that are relatively prime to n.Let's look at
phi(p^2). There are p^2 natural numbersup to and including
p^2. How does a natural number getto be relatively prime to
p^2? Well, since p^2 has onlythe prime divisor p, a number
fails to be relatively primeto p^2 if it's a multiple of p.
Everything else is thenrelatively prime, and enters into the
count to evaluatephi(p^2).So, the question becomes this: how
many multiples of pare there in the set{1, 2, ..., p, p+1,
..., 2p, 2p+1, ...3p, ..., 4p, ..., p^2}Hint: I've indicated
the first 4 of them, as well as thefinal one. Can you spot the
pattern? Can you then findout how many numbers in {1, ..., p^2}
fit that pattern?Having done that, you have the numbers {1,
..., p^2} a setwith p^2 elements. You have just generated
_______ membersthat must be excluded since they're not
relatively prime top^2 (fill in the blank yourself). How would
you determinehow many are left?That's it.Dale===
===
Subject:
: Re:
pi, 2 pi and frivolity>To me it seems rather curious that the
human race has chosen pi ratherthan>2 pi as the important
number that deserves its own special symbol.>So much so that
you're not the first one to have thought about it. Read>23 no.
3, pp. 7-8 (2001)).> I remember Serge Lang saying much the same
thing in a talk he gave> 20 or 30 years
ago.here:http://www.math.utah.edu/~palais/pi.htmlHe puts his
case well, says everything I did, and more.===
===
Subject:
: Re: pi,
2 pi and frivolityI was actually thinking of tau for 2 pi,
because then pi would be tau on twolegs :-). If you're right
though I'd be curious why he chose pi/2, itdoesn't seem any
better choice than or pi or 2 pi to me.===
===
Subject:
: Re: pi, 2
pi and frivolity> What's the point? The important thing is
that everyone agrees on the same> symbols. Using your
arguments, why are there sixty minutes in an hour?The point is
that maths would have been more convenient and more intuitiveif
2 pi, rather than pi, had been chosen as the number with its
own specialsymbol.I agree that having everyone using the same
symbol is an important point.> Using your arguments, why are
there sixty minutes in an hour?There are 60 minutes in an hour
for historical reasons, which is exactly thesame reason that I
am claiming we use pi and not 2 pi.> a formula, sometimes just
pi. How do you judge the fundamentality of anumber?I judge how
fundamental a number is by whether it appears in
fundamentalformulas and equations. I judge how fundamental an
equation is by how oftenit is quoted in deriving other
equations. A similar judgement is how earlyit appears in a
course of study in maths.Geometrically speaking, to me C = 2
pi r is the fundamental equationinvolving pi, it appears with
2, so 2 pi is the fundamental number. Ibelieve that C = 2 pi r
is the fundamental equation because I imagine thatit is
substituted in more proofs that any other formula involving
pi, andbecause its surely the first one involving pi presented
to a geometrystudent in any reasonable course of study, except
possibly C = pi D. Iregard r as a more fundamental measurement
of a circle than D, because r andnot D is used to define a
circle, so I discard C = pi D as being
relativelyunimportant.Analytically speaking, to me the most
fundamental use of pi is the period ofthe periodic functions
cos and sin, and that period is 2 pi.Using 2 pi does make
certain formulas easier to remember, in my opinion, butwe can
be more objective that that. Simply count the number of
mathematicaloperations in the formula. You've got an extra 2
there, that's one moreoperation.===
===
Subject:
: Re: pi, 2 pi and
frivolitythe area of the *sphere* doesn't use a factor of
2(the area of the circumferential circle is a quarterof the
sphere's area .-)> because (2 r) = d. The circumference and
area both use pi, neither uses 2
pi.http://www.channel1.com/users/bobwb/synergetics/photos/
x6girdle6.html--Give Earth a Trickier Dick Cheeny -- out of
office, after GIGA
years.http://www.benfranklinbooks.com/http://www.rand.org/
publications/randreview/issues/rr.12.00/http://
members.tripod.com/~american_almanac===
===
Subject:
: Re: pi, 2 pi
and frivolity <87ptb97v9b.fsf@phiwumbda.org>
<aWo6c.16714$uh.12190@fed1read02 Discussion, linux)>> What's
the point? The important thing is that everyone agrees on the
same>> symbols. Using your arguments, why are there sixty
minutes in an hour?>> Surely, 100 minutes/hour would be easier
to deal with.> If you had 100 seconds in a minute, 100 minutes
in an hour, and 10 hours in> a day, the second wouldn't have
to change by all that much.Please watch the attributions. You
have an attribution to me upthere, but you don't quote
anything I said.-- What I've learned is that [mathematicians
are] the gatekeepers, andseem to have almost absolute power
when it comes to mathematics. -- James Harris, on All I Really
Ever Needed to Know I Learned in /Ghostbusters/.===
===
Subject:
: Re:
pi, 2 pi and frivolity <87smg57vdj.fsf@phiwumbda.org
<3tck509fisn2tqe83qsgtdq9itr8ed8thf@4ax.com Discussion,
linux)> I don't know anything about Albert Eagle or his book,
and I> don't know what's mentioned in Dudley. But I do know >
something about Fourier series and transforms, and I> don't
see anything particularly crankish about the quote.Sorry, I
didn't mean to imply that Eagle was a crank or that the
quotewas crankish.I have vague memory that Eagle's quote was
just mentioned in the crankbook in passing, not that Dudley
called him a crank. In any case, Arturo says Eagle doesn't
appear in the index and so this vague memoryis probably wrong.
I probably saw the quote elsewhere.-- If you *still* believe
that [my proof is wrong], then I have to thinkthat your mind
is limited [...], and it may be the case that noteveryone
*can* achieve that, as the mental wiring may not be there
forthe task. -- James Harris, on faculties needed to accept
his proof.===
===
Subject:
: Re: pi, 2 pi and frivolityOriginator:
hack@watson.ibm.com (hack)>I have vague memory that Eagle's
quote was just mentioned in the crank>book in passing, not
that Dudley called him a crank. In any case, >Arturo says
Eagle doesn't appear in the index and so this vague memory>is
probably wrong. I probably saw the quote elsewhere.I saw the
quote in Imre Latakos' Proofs and Refutations. I don't havethe
book here, so I can't verify whom it was attributed
to.Michel.===
===
Subject:
: Re: pi, 2 pi and frivolity
<3tck509fisn2tqe83qsgtdq9itr8ed8thf@4ax.com
<87n06d72dl.fsf@phiwumbda.org> <c3fahc$rfs$2@news.btv.ibm.com
Discussion, linux)>>I have vague memory that Eagle's quote was
just mentioned in the crank>>book in passing, not that Dudley
called him a crank. In any case, >>Arturo says Eagle doesn't
appear in the index and so this vague memory>>is probably
wrong. I probably saw the quote elsewhere.> I saw the quote in
Imre Latakos' Proofs and Refutations. I don't have> the book
here, so I can't verify whom it was attributed to.I have that
book, and I've read parts of it, so that might be where Isaw
it. -- And a journal can beg me for the right to publish it
[...] becauseI'd rather see it in People magazine [...]
--James Harris on his simple proof of Fermat's last
theorem===
===
Subject:
: Re: approximately equal to> Why is there a
almost equal to and approximately equal to what is> the
difference?> se unicode Mathematical Operators>
http://www.unicode.org/charts/PDF/U2200.pdfMisnomer. Neither
almost equal to nor approximately equal to aremathematical
operators.===
===
Subject:
: Re: approximately equal toalmost equal
to has below it, asymptotic to which has, to me, a meaning
suggesting the notation should be used when there is a
parameterq and a limit k such that lim q->k (f-g)=0 means f is
almost equal to g. Or something similar.>>Why is there a almost
equal to and approximately equal to what is>>the
difference?>>se unicode Mathematical
Operators>>http://www.unicode.org/charts/PDF/U2200.pdf>
Misnomer. Neither almost equal to nor approximately equal to
are> mathematical operators.The unicode people disagree with
you, apparently.===
===
Subject:
: A Number Theory QuestionHelloI
would like to find a concise description of the logical
foundationsof number theory. I'm not particularly skilled or
knowledgeable inmath but I think I'm fairly good at
understanding logical arguments.What I'm looking for is the
most basic step... how does it start.Appearance of 1 & zero
and the general boot-strap process from whichthe rest
appears.more or less like the following. I have phrased it in
a sort ofrelativistic way since I believe that Einstein's
insights of thenecessity of describing things in terms of a
frame of reference arevalid for all logical systems.Rough
overview: If I understand correctly... first a non-zero value
isproved (or observed), then zero is inferred through
implication, thena negative value is instantiated (implied,
not disprovable...), andso on. Where does quantization,
definition of whole numbers (e.g. 1,0, -1, 2, -2, etc.)
appear?Does it go something like this?First, something
different is noticed. This would be, if I understandcorrectly,
an instantiation of NOT EQUAL. I think that this is thesimplest
logical condition that could initiate the necessary chain
ofinferences.Second, the observation of the instantiation of
NOT EQUAL implies areference value... something to which the
observed quantity is notequal. Instantiation of NOT EQUAL
requires two quantities and aframework (logical system or
something) within which the values can beobserved as
different.This implies zero. The reference frame for all
observations in alllogical systems (number lines, electrical
voltage, velocity, etc.) isalways the local reference value or
zero value.Third, existence of two values implies an
alternative reference framefrom which the value previously
designated as non-zero is the localreference, or zero. These
two reference frames or differentperspectives on these two
non-equal quantities provide a logicalframework to define
polarity or direction or GREATER THAN and LESSTHAN... that is,
vector comparisons.choose A>0 or A<0.Presume one's reference is
A>0. This implies a logical structure ?<0or B<0. Any proof that
B doesn't exist will fail. Any logicalstructure dependent upon
the existence of B such that B<0 will bevalid. B is therefore
instantiated.One then has three quantities, B<0<A. This
bootstrap process wouldcontinue ad infinitum.That seems
straight-forward.How does quantization occur? A sort of
extension of Occam's Razor...the least information required to
describe a value satisfying therequirements of A is a single
bit?What minimal set of logical (mathematical) operators are
necessary toquantize values into whole numbers?Michael
McGinnismike@michael-mcginnis.comCosmic Laws:1) Rust never
sleeps2) Appearance is Reality3) These Laws are subject to
change without notice===
===
Subject:
: Re: A Number Theory
Question>I would like to find a concise description of the
logical foundations>of number theory. I'm not particularly
skilled or knowledgeable in>math but I think I'm fairly good
at understanding logical arguments.Books on metamathematics
will describe this. Try the newsgroups sci.mathor perhaps
sci.math.research. Hans Aberg===
===
Subject:
: Re: A Number Theory
Question> Hello> I would like to find a concise description of
the logical foundations> of number theory. I'm not particularly
skilled or knowledgeable in> math but I think I'm fairly good
at understanding logical arguments. [ deleted ]You might want
to consult Frege's theorem. The following URL will getyou
started: http://plato.stanford.edu/entries/frege-logic/--
Julian V. NobleProfessor Emeritus of
Physicsjvn@lessspam.virginia.edu
^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ For there was
never yet philosopher that could endure the toothache
patiently. -- Wm. Shakespeare, Much Ado about Nothing. Act v.
Sc. 1.===
===
Subject:
: Re: A Number Theory Question> Hello> I would
like to find a concise description of the logical foundations>
of number theory. I'm not particularly skilled or
knowledgeable in> math but I think I'm fairly good at
understanding logical arguments.> What I'm looking for is the
most basic step... how does it start.> Appearance of 1 & zero
and the general boot-strap process from which> the rest
appears.There are many different ways to define the logical
foundations. Oneof the simplest most self contained ones are
the PeanoAxiom's. Quoting fromEric W. Weisstein.
http://mathworld.wolfram.com/PeanosAxioms.html Foundations of
MathematicsAxiomsPeano's Axioms 1. Zero is a number.2. If a is
a number, the successor of a is a number.3. zero is not the
successor of a number.4. Two numbers of which the successors
are equal are themselves equal.5. (induction axiom.) If a set
S of numbers contains zero and also the successor of every
number in S, then every number is in S.Peano's axioms are the
basis for the version of number theory known asPeano
arithmetic.> more or less like the following. I have phrased
it in a sort ofrelativistic way since I believe that
Einstein's insights of the> necessity of describing things in
terms of a frame of reference are> valid for all logical
systems.> Rough overview: If I understand correctly... first a
non-zero value is> proved (or observed), then zero is inferred
through implication, then> a negative value is instantiated
(implied, not disprovable...), and> so on. Where does
quantization, definition of whole numbers (e.g. 1,> 0, -1, 2,
-2, etc.) appear?> Does it go something like this?> First,
something different is noticed. This would be, if I
understand> correctly, an instantiation of NOT EQUAL. I think
that this is the> simplest logical condition that could
initiate the necessary chain of> inferences.> Second, the
observation of the instantiation of NOT EQUAL implies a>
reference value... something to which the observed quantity is
not> equal. Instantiation of NOT EQUAL requires two quantities
and a> framework (logical system or something) within which
the values can be> observed as different.> This implies zero.
The reference frame for all observations in all> logical
systems (number lines, electrical voltage, velocity, etc.) is>
always the local reference value or zero value.> Third,
existence of two values implies an alternative reference
frame> from which the value previously designated as non-zero
is the local> reference, or zero. These two reference frames
or different> perspectives on these two non-equal quantities
provide a logical> framework to define polarity or direction
or GREATER THAN and LESS> THAN... that is, vector
comparisons.> choose A>0 or A<0.> Presume one's reference is
A>0. This implies a logical structure ?<0> or B<0. Any proof
that B doesn't exist will fail. Any logical> structure
dependent upon the existence of B such that B<0 will be>
valid. B is therefore instantiated.> One then has three
quantities, B<0<A. This bootstrap process would> continue ad
infinitum.> That seems straight-forward.> How does
quantization occur? A sort of extension of Occam's Razor...>
the least information required to describe a value satisfying
the> requirements of A is a single bit?> What minimal set of
logical (mathematical) operators are necessary to> quantize
values into whole numbers?> Michael McGinnis>
mike@michael-mcginnis.com> Cosmic Laws:> 1) Rust never sleeps>
2) Appearance is Reality> 3) These Laws are subject to change
without notice-- Use of tools distinguishes Man from Beast.
And UNIX users from WINDOZE lusers.===
===
Subject:
: Re: some
questions about Hensel lifting[...]>With nowadays heavy use of
asymptotically fast integer multiplication>and univariate
polynomial algebra, this is no longer the case:>quadratic
lifting performed in a reasonable way is, to my knowledge,>not
beatable. An example of a reasonable way would be the tree
lift>approach due to Victor Shoup, presented on pp. 424-425 of
the>above-cited text. A very different approach is based on
Groebner bases>over the integers. It is also a quadratic lift
and tends to compete>well in terms of speed. It can suffer
from a bad prime problem but>in practice that is not an
issue.Do you have a reference on the Groebner bases approach
to factorization ? I've designed such an algorithm on my own
for meditor, but the efficiency is very deceiving so far. I
need clues to what I'm doing wrong.Raphael===
===
Subject:
: Re: some
questions about Hensel lifting> [...]>With nowadays heavy use
of asymptotically fast integer multiplication>and univariate
polynomial algebra, this is no longer the case:>quadratic
lifting performed in a reasonable way is, to my knowledge,>not
beatable. An example of a reasonable way would be the tree
lift>approach due to Victor Shoup, presented on pp. 424-425 of
the>above-cited text. A very different approach is based on
Groebner bases>over the integers. It is also a quadratic lift
and tends to compete>well in terms of speed. It can suffer
from a bad prime problem but>in practice that is not an
issue.> Do you have a reference on the Groebner bases approach
to factorization ? I've designed such an algorithm on my own
for meditor, but the efficiency is very deceiving so far. I
need clues to what I'm doing wrong.> RaphaelI apologize for
the length of this reply; most is tied in an example,with a
largish input reproduced in case anyone else wants to work
withit.The method is discussed in:[D. L.] Revisiting strong
Groebner bases over Euclidean domains.Submitted.If you want
all the details, a postscript copy may be found
athttp://download.wolfram.com/?key=THVBL1As the title would
indicate, the paper is really about computingGroebner bases
over Euclidean domains. Some special cases present verynice
(in my opinion) techniques for doing things not
ordinarilyassociated with such bases, and I discuss some of
these. Polynomialfactorization appears in a couple of the
example sections. I'll showthe code I use below so you don't
need to download the paper if youonly want to see code.I
should remark that the ways in which Groebner bases might
apply topolynomial factorization appear to depend on the
domain over which onefactors. For the common case of
univariate polynomials over therationals, to my knowledge it
can only be used for Hensel lifting, andcannot do other parts
of the task. Other domains, e.g. bivariatepolynomials over
finite fields, are amenable to use of Groebner basesto do more
than just lift factors (some aspects of this are covered inthe
reference). Unlike the Hensel lifting case, I do not believe
thesemethods to be particularly fast, but they may be at least
reasonable.As for univariate Hensel lifting, the example I use
in the reference,and below, is fromM. van Hoeij (2002).
Factoring polynomials and the knapsack problem.Journal of
Number Theory 95:167-181.We begin by constructing a
-polynimial of degree 190, whose roots areall sums of roots of
a degree 20 polynomial.poly1 = x^20 - 5*x^18 + 864*x^15 -
375*x^14 - 2160*x^13 + 1875*x^12 + 10800*x^11 + 186624*x^10 -
54000*x^9 + 46875*x^8 + 270000*x^7 - 234375*x^6 - 2700000*x^5
- 1953125*x^2 + 9765625; rts = x /. Solve[poly1 == 0, x]; sums
= Flatten[Table[rts[[i]] + rts[[j]], {i, 19}, {j, i + 1, 20}]];
newpoly = Expand[Times @@ (x - N[sums, 200])]; newpoly =
Chop[newpoly] /. a_Real -> Round[a]; In case you want to work
with this polynomial and do not havemathematica and do not
want to rewrite the construction code above,here it is.In[6]:=
InputForm[newpoly]Out[6]//InputForm=
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1464579659652576451900553947008107195467824*x^130 +
1859494487344549140946539844665219526596000*x^131 +
282768363743611451141308008361959136260000*x^132 -
99076965299075976715488344501554246306080*x^133 -
19368832387894079515801483713647563485000*x^134 +
2575625130823917726899166123153184785696*x^135 +
1002008946355555719764560762727277776200*x^136 +
64600551006156546125329046134701900000*x^137 -
5945959355037571095414802632914940000*x^139 +
1917527740657775445746608759689040116*x^140 +
387615087856766742510221578840284000*x^141 -
39909333064565109277718890825196250*x^142 -
29148561313173475905090517885634400*x^143 +
3237316432618038080433617702952000*x^144 +
1412437877293501891469764644531552*x^145
-179199253293377468610099823319700* x^146 -
47165849594104231690867916100000*x^147 +
3306888055787944054538380925880*x^148
+2195001449642924500817837304000* x^149 +
109828724220903706832512368252*x^150 -
87653534939706490799859727200*x^151
-11579942607226771890474337500*x^152 +
2775750991804671182260443360*x^153
+566297285176056600366032125*x^154 -
44647572359536813529921088*x^155 -
30206991981195282769112250*x^156 +
1384009997987828815380000*x^157 +
1651039293752957752389765*x^158 -
216155508822073386540000*x^159 - 57162249087627923376168*x^160
+ 11017258598962782590400*x^161 + 1462564207283754026250*x^162
- 259055867794780032480*x^163 - 58056443308895422500*x^164 +
1629340768614264192*x^165 + 2175914230432545750*x^166 +
117526456730520000*x^167 - 61161667283546100*x^168 -
6350251415004000*x^169 + 816272603315991*x^170
+452956642790400*x^171 - 53873241913750*x^172 -
26824309004160*x^173 + 6025540138875*x^174 +
820004125536*x^175 - 315373944600*x^176 - 12828240000*x^177 +
9522029130*x^178 + 33912000*x^179 - 142192764*x^180 +
8640000*x^181 + 1428750*x^182 - 276480*x^183 - 97500*x^184 +
3456*x^185 + 3825*x^186- 90*x^188 + x^190We will first factor
modulo the 4000th prime.mod = Prime[4000];fax =
FactorList[newpoly,Modulus[Rule]mod];fax = First/@Rest[fax];It
turns out that all factors are of degree 5. Below I show the
firstand last obtained by the factorization above.In[11]:=
InputForm[{First[fax],Last[fax]}]Out[11]//InputForm= {29129 +
11572*x + 26850*x^2 + 6066*x^3 + 7628*x^4 + x^5, 2223 +
32743*x + 26837*x^2 + 36575*x^3 + 37682*x^4 + x^5}Here is the
Groebner basis code for Hensel
lifting.liftfactors[fax_,poly_,mod_,pow_]:= Module[
{modpow=mod,top=Ceiling[Log[2,pow]],liftedfax=fax},
Do[modpow=If[j[Equal]top,mod^pow,modpow^2];
liftedfax=Expand[liftedfax^2,Modulus[Rule]modpow]; liftedfax=
Map[Last[ GroebnerBasis[{modpow,poly,#},
CoefficientDomain[Rule]Integers]]&,liftedfax], {j,top} ];
liftedfax ]We lift to mod^36, which turns out to be about
where you want to go toemploy a factor recombination e.g. van
Hoeij's knapsack method.In[13]:=
Timing[liftedfax=liftfactors[fax,newpoly,mod,36];]Out[13]=
{5.4 Second, Null}Here we see what thos first and last factors
have become.In[14]:=
InputForm[{First[liftedfax],Last[liftedfax]}]Out[14]//
InputForm=
{
680487749454858952282038502387372022528970358812327019687511394
950354171657599283610462704984973893197803392120692249935909824
6798111720633049841165246681818354571 +
140785165961937329820230724246794571497670834312223452693267207
552784833807630197850397182708504601462171386784925358190426612
521242434588755985813380473210521328733*x
+
291774259227280294326710779691745216853199372635298156971701060
270718202449413083089576332824740467891620836924246016103742658
044899164880452804642531368833966847354*x^2
+
163364772006444454490473772173788194511912512341237692890087683
317755858816439358377839313400316431935133825774563069062387180
32412244472567168212500050297152333551*x^3 +
119621772481414777919574138735616856376535827148496347477148361
875978117000469524557413325436469443039232253146219150287338336
140650282044956422244103674731827413489*x^4 +
x^5,-
137168588684633482123617495348737010698815212010665165126013607
477918259119541666859466132192164865435665386518645298052613151
710926208310139801223822564327270759680
+
162193328832167013877551713543505192276041395003058147804698245
976972518905711188037104211843059777856765846950903394237023419
142345698754523470945305047761510587191*x -
424888619563027153928414203053423309757925975416818829749681593
242343551221035217725534976177709599663145858306776779991614924
96979642455156577028827788398064447283*x^2 +
437195438598294657668090927055125889414319834344105801981577435
154477071738831659658111312293451262626015248009286959961014008
06496072409866008007394882048095931873*x^3
+
209079256550525904019872482267786533119380319417078881064096857
320068695030052148008596561391897435671500178195445666120319621
571319337521886985595313842735819283264*x^4 + x^5}I'll note
that dedicated univariate polynomial manipulation code,
e.g.using dense lists, will do this substantially faster (we
get a factorof 5 or so for this particular example). I show
the slower code aboveso it is transparent how the method was
has its roots in Groebnerbases over the integers.iel
LichtblauWolfram Research===
===
Subject:
: Re: some questions about
Hensel lifting>[...]>>With nowadays heavy use of
asymptotically fast integer multiplication>>and univariate
polynomial algebra, this is no longer the case:>>quadratic
lifting performed in a reasonable way is, to my
knowledge,>>not beatable. An example of a reasonable way would
be the tree lift>>approach due to Victor Shoup, presented on
pp. 424-425 of the>>above-cited text. A very different
approach is based on Groebner bases>>over the integers. It is
also a quadratic lift and tends to compete>>well in terms of
speed. It can suffer from a bad prime problem but>>in practice
that is not an issue.>Do you have a reference on the Groebner
bases approach to factorization ? I've designed such an
algorithm on my own for meditor, but the efficiency is very
deceiving so far. I need clues to what I'm doing
wrong.>Raphael> [...]> The method is discussed in:> [D. L.]
Revisiting strong Groebner bases over Euclidean domains.>
Submitted.> If you want all the details, a postscript copy may
be found at> http://download.wolfram.com/?key=THVBL1> [...]I've
been told that the postscript download from that link can
beproblematic for some browsers. I have put a pdf version
athttp://download.wolfram.com/?key=4LV1EMIt is only about
180K, in contrast to the roughly 3.2M size of thepostscript
version.iel LichtblauWolfram Research===
===
Subject:
: Re:
Recommendation between two programs...Have you considered
Maxima, which is GPL'd?http://maxima.sourceforge.net/> I am
looking at getting a math software program for school. I am>
currently a college student, and I was just introduced to
Maple 9 by> my Calculus instructor. WOW! I never knew such a
product existed.> Thinking I would do a little more research
on Maple 9, I actually> discovered Mathematica, IDL, and
Matlab (actually I was introduced to> Matlab as well, but we
didnt tool around with it).> Here is a little about me: I am
working toward a degree in> Biochemistry; I want to take as
many forms of higher maths as I can> (Calc II, III, etc.); I
have a little computer knowledge; price> doesnt matter too
much (Im not rich but I could justify spending> about two
hundred dollars if I could find a neat and useful software>
application); my only other math tools that Ive used were a
TI-83+ SE> (and a slide rule for fun). I am looking at buying
this type of> software as an aide in my studies and to double
check my work.> Right now, I am looking at either Mathematica
or Maple 9. Any help or> advice you offer would be greatly
appreciated.> Styron from North Carolina===
===
Subject:
: Re:
Journals and my papersDear newsgroup:The Cambridge University
informed me that the paper on non-solvablequintic is too
computational for the proceeding, thus I need to find
asuitable place.The Mediterranean Journal of Math will send
the papers on Bessel forreferee consideration.I have a lot of
papers on Riccati and two papers on new methods wasrejected
today.I have full confidence about the methods and that it was
not clear forthe referee what I was talking about, although I
am not charging thatthe referee was necessarily not
insightful.The papers were designed for readers to learn the
techniques notproviding a complicated solution particularly
with comparison with thesoftware.The report is enclosed. There
were two referees one declined toreport. I will send the papers
for more suitable places. Eventually Ineed to find a suitable
community responding to these kind of researchfollowed by my
own lecture notes.SincerelyDr.Mehran
Basti---------------------------------------------------------
--------------------Referee's ReportToElectronic Journal of
Differential EquationsNew methods for solving Riccati
differential equationsNew methods for solving Riccati
differential equations IIIt is the opinion of the referee that
these two papers contributenothing to the mathematical
literature.Their theoretical content is nil and they are
completely unsuitablefor publication in mathematics
journal.The formal manipulations that the author offers seem
to be moving inprecisely the wrong direction (converting
linear problems intononlinear ones only special cases of which
can be solved) and this canhardly be considered the start of a
new era in mathematics. Thenarrative is vague and at time
incoherent. I recommend a clear andfirm rejection.===
===
Subject:
:
Re: Journals and my papers> Referee's Report> To> Electronic
Journal of Differential Equations> New methods for solving
Riccati differential equations> New methods for solving
Riccati differential equations II> It is the opinion of the
referee that these two papers contribute> nothing to the
mathematical literature.> Their theoretical content is nil and
they are completely unsuitable> for publication in mathematics
journal.> The formal manipulations that the author offers seem
to be moving in> precisely the wrong direction (converting
linear problems into> nonlinear ones only special cases of
which can be solved) and this can> hardly be considered the
start of a new era in mathematics. The> narrative is vague and
at time incoherent. I recommend a clear and> firm
rejection.Nothing new here. Referees don't always put sincere
effort intounderstanding the subject. Especially if the paper
is written badly. If thepaper is written badly plus it has new
era math announcements, the refereehave all the reasons in the
world to suspect that it's a lunacy. And, then,when your
reputation is destroyed, how can he seriously concider yet
anothersubmission from the same author? So consider this as a
game: youartificially handicap yourself by giving unfair
advantage to your opponent.Remove those comments asap.